text stringlengths 1 2.12k | source dict |
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then
$\begin{array}{ccc}{a}_{11}B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]& {a}_{12}B=\left[\begin{array}{cc}10& 12\\ 14& 16\end{array}\right]& C=\left[\begin{array}{cc}{a}_{11}B& {a}_{12}B\\ {a}_{21}B& {a}_{22}B\end{array}\right]=\left[\begin{array}{cccc}5& 6& 10& 12\\ 7& 8& 14& 16\\ 15& 18& 20& 24\\ 21& 24& 28& 32\end{array}\right]\end{array}$
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# What is the metric tensor on the n-sphere (hypersphere)?
I am considering the unit sphere (but an extension to one of radius $r$ would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and $n-1$ angular coordinates) would be preferable.
I know that on the 2-sphere we have $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$ (in spherical coordinates) but I'm not sure how this generalizes to $n$ dimensions.
Added note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful.
I will define the metric of $$S^{n-1}$$ via pullback of the Euclidean metric on $${\mathbb{R}}^{n}$$.
To start with we take $$n$$-dimension Cartesian co-ordinates: $$(x_1,x_2......x_n)$$. The metric here is $$g_{ij }= \delta_{ij}$$, where $$δ$$ is the Kronecker delta.
We specify the surface patches of $$S^{n-1}$$ by the parametrization $$f$$: $$x_1=r{\cos{\phi_1}},$$
$$x_p=r{\cos{\phi_p}}{\Pi_{m=1}^{p-1}}{\sin{\phi_{m}}},$$
$$x_n=r{\prod_{m=1}^{n-1}}{\sin{\phi_{m}}},$$
Where $$r$$ is the radius of the hypersphere and the angles have the usual range.
We see that the pullback of the Euclidean metric $$g'_{ab} = (f^*g)_{ab}$$ is the metric tensor of the hypersphere. Its components are:
$$g'_{ab} = g_{ij} {\frac{\partial{x_i}}{\partial{\phi_a}}} {\frac{\partial{x_j}}{\partial{\phi_b}}} = {\frac{\partial{x_i}}{\partial{\phi_a}}}{\frac{\partial{x_i}}{\partial{\phi_b}}}$$
We get $$2$$ cases here:
i) $$a>b$$ or $$b>a$$, For these components one obtains a series of terms with alternating signs which vanishes, $$g'_{ab}=0$$ and thus all off-diagonal components of the tensor vanish.
ii) $$a=b$$,
$$g'_{11}=1$$
$$g'_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$
where $$2\leq a\leq {n-1}$$
The determinant is very straightforward to calculate:
$$\det{(g'_{ab})} = {r^2} \prod_{m=1}^{n-1} g'_{mm}$$
Finally, we can write the metric of the hypersphere as: | {
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Finally, we can write the metric of the hypersphere as:
$$g' = {r^2} \, d\phi_{1}\otimes d\phi_{1} + {r^2} \sum_{a=2}^{n-1} \left( \prod_{m=1}^{a-1} \sin^2{\phi_{m}} \right) d\phi_{a} \otimes d\phi_{a}$$ | {
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• Great, thanks for the step-by-step explanation. Although what you've written is fairly clear, you can make it clearer by using the TeX environment, with the $ delimiters, like instead of g'rϕk, write $g'_{r\phi_k}$, which comes out as$g'_{r\phi_k}$. May 27, 2013 at 11:39 • This is a very good answer and very useful too, as it is not that easy to find this piece of information on the net. Proper formatting would make it great. Oct 20, 2014 at 9:12 • @GiuseppeNegro Done. Thanks for the feedback. Nov 10, 2014 at 14:04 • It seems$g'_{11}=r^2$, no? Nov 13, 2018 at 14:17 • Can somebody explain more thoroughly why the off diagonal terms vanish? Apr 16, 2019 at 11:15$\newcommand{\Reals}{\mathbf{R}}$For posterity: Fix$r > 0$, and let$S^{n}(r)$denote the sphere of radius$r$centered at the origin in$\Reals^{n+1}$. Stereographic projection from the north pole$N = (0, \dots, 0, 1)$on the unit sphere$S^{n} = S^{n}(1)$defines a diffeomorphism$\Pi_{N}:S^{n} \setminus \{N\} \to \Reals^{n}given in Cartesian coordinates by \begin{align*} \Pi_{N}(x_{1}, \dots, x_{n}, x_{n+1}) &= \frac{1}{1 - x_{n+1}}(x_{1}, \dots, x_{n}), \\ \Pi_{N}^{-1}(t_{1}, \dots, t_{n}) &= \frac{(2t_{1}, \dots, 2t_{n}, \|t\|^{2} - 1)}{\|t\|^{2} + 1}. \end{align*} In these coordinates, the induced (round) metric on the unit sphere is well-known (and easily checked) to be conformally-Euclidean: $$g(t) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + 1)^{2}}.$$ Stereographic projection from the north pole(0, \dots, 0, r)$of$S^{n}(r)$is given by the scaled mapping$x \mapsto t = r\Pi_{N}(x/r)$, whose inverse is$t \mapsto x = r\Pi_{N}^{-1}(t/r)\$, i.e., \begin{align*} r\Pi_{N}(x_{1}/r, \dots, x_{n}/r, x_{n+1}/r) &= \frac{1}{r - x_{n+1}}(x_{1}, \dots, x_{n}), \\ r\Pi_{N}^{-1}(t_{1}/r, \dots, t_{n}/r) &= \frac{\bigl(2t_{1}, \dots, 2t_{n}, r(\|t/r\|^{2} - 1)\bigr)}{\|t/r\|^{2} + 1}. \end{align*} The induced metric in these coordinates is consequently $$r^{2} g(t/r) = \frac{4 (dt_{1}^{2} + \cdots + | {
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induced metric in these coordinates is consequently $$r^{2} g(t/r) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t/r\|^{2} + 1)^{2}} = \frac{4r^{4} (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + r^{2})^{2}}.$$ | {
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# A problem in understanding the Intermediate Value Theorem
So, IVT essentially says if a function $$f$$ is continuous over an interval $$[a,b]$$, then the function $$f$$ will take up all the values between $$f(a)$$ and $$f(b)$$ at least once at some point in the interval $$[a,b]$$.
Now, suppose $$f$$ is continuous between $$[3,7]$$ and $$f(3) = 4$$ and $$f(7) = 25$$
According to IVT, $$f(3) = 4 < m < 25 = f(7)$$, i.e., the function takes up all the values between $$f(3)$$ and $$f(7)$$ at least once in the interval. But, according to the below picture, clearly, the function within that interval takes up values different that do not satisfy the above inequality. Please help me fill the gap in my understanding of IVT here.
Please do correct me if I'm mistaken.
This doubt arose from the below KhanAcademy question | {
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Please do correct me if I'm mistaken.
This doubt arose from the below KhanAcademy question
• IVT doesn't tell you that ALL values must be between $f(a)$ and $f(b)$ -- it tells you that every value between $f(a)$ and $f(b)$ must be attained. – Nick Peterson Feb 25 at 6:59
• The IVT says that $f$ reaches all values between $f(3)$ and $f(7)$. It doesn't say it can't reach other values as well. – Mathematician 42 Feb 25 at 6:59
• @SathvikR. If it helps: you seem to be interpreting it as: $x\in(a,b)$ implies $f(x)\in (f(a), f(b))$ (assuming $f(a)<f(b)$, etc). But the actual implication is: for all $y\in (f(a), f(b))$, there exists $x\in(a, b)$ so that $f(x)=y$. – Nick Peterson Feb 25 at 7:06
• IVT only says that all values between $4$ and $25$ will be obtained. And in that picture they clearly all are. The IVT says NOTHING at all about values outside $[4,25]$. It doesn't say they are obtained. It doesn't say they aren't obtained. – fleablood Feb 25 at 7:21
• Okay.... it says that every possible output value between $f(3) = 4$ and $f(7) = 25$ will be obtained somewhere in the interval between $3$ and $7$. But it doesn't say that values between $4$ and $25$ are the only values obtained. If I told you that every dog in the Golden Gate Dog shelter has an owner who lives in San Francisco, that does not mean everyone who lives in San Francisco owns a dog in the Golden Gate Dog shelter.... to be continued.... – fleablood Feb 25 at 7:29
The intermediate value theorem tells you that, in that context, for each $$y\in[4,25]$$, there is some $$x\in[3,4]$$ such that $$f(x)=y$$. It does not tell you that if $$x\in[3,4]$$, then $$f(x)\in[4,25]$$. So, there is no contradiction if $$f\bigl([3,4]\bigr)$$ contains values outside $$[4,25]$$. | {
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# Thread: Help with this word problem!
1. ## Help with this word problem!
The value of a new car of a particular make, A, decreases by 22.5% during the first year. In the second year, its value decreases by 12.5% of its value at the beginning of the year. The value of a new car of another make, B, decreases by 25% during the first year. In the second year, its value decreases by 15% of its value at the beginning of the year. A new car of make A costs $48,000 and a new car of make B costs$42,000.
(a) What is the difference between the values of the two cars at the end of the firts year? Express this differnce as a percentage of the value of the car of make A at the end of the first year.
(b) Calculate the difference between the values of the two cars at the end of the second year.
Thankyou!
2. Originally Posted by laser2302
The value of a new car of a particular make, A, decreases by 22.5% during the first year. In the second year, its value decreases by 12.5% of its value at the beginning of the year. The value of a new car of another make, B, decreases by 25% during the first year. In the second year, its value decreases by 15% of its value at the beginning of the year. A new car of make A costs $48,000 and a new car of make B costs$42,000.
(a) What is the difference between the values of the two cars at the end of the firts year? Express this differnce as a percentage of the value of the car of make A at the end of the first year.
(b) Calculate the difference between the values of the two cars at the end of the second year.
Thankyou!
If something decreases by 22.5% means it is worth 77.5% perfect of the original price.
In the beggining the car was worth $x$ after the first year is it $.775x$ after the second year it decreases by 12.5% meaning its value is 87.5% of the new price thus,
$.875(.775x)=.678125x$
Similarily the second car after two years is,
$.85(.75x)=.6375x$ and $.75x$ after one year. | {
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The first problem asks for the difference after the first year.
The first car is worth $.775\times 48,000=37,200$. The second car is worth $.75\times 42,000=31,500$ The difference is,
$37,200-31,500=5,700$
After two years, you have the new values,
$.875\times 37,200=32,250$
$.85\times 31,500=26,775$
Their difference is,
$32,250-26,775= 5475$
3. Originally Posted by ThePerfectHacker
The first problem asks for the difference after the first year.
The first car is worth $.775\times 48,000=37,200$. The second car is worth $.75\times 42,000=31,500$ The difference is,
$37,200-31,500=5,700$
The question asks to write this out in terms of the percentage of the price of car A, so you divide 5,700 by the price of car A
$\frac{5700}{37200}=0.153225806\approx\boxed{15
\text
{percent}}$
4. Hello, laser2302!
Are you sure you can't work this out yourself?
[quote]The value of a new car of a particular make, $A$, decreases by 22.5% during the first year.
In the second year, its value decreases by 12.5% of its value at the beginning of the year.
The value of a new car of another make, $B$, decreases by 25% during the first year.
In the second year, its value decreases by 15% of its value at the beginning of the year.
A new car of make $A$ costs $48,000 and a new car of make $B$ costs$42,000.
(a) What is the difference between the values of the two cars at the end of the first year?
Express this differnce as a percentage of the value of the car of make A at the end of the first year.
(b) Calculate the difference between the values of the two cars at the end of the second year.
During year 1, make $A$ loses $22.5\%\text{ of }\48,000 \:=\:0.225 \times 48,000 \:=\:\10,800$
. . . At the end of year 1, it is worth only: $\48,000 - 10,800 \:= \:\37,200$
During the year 2 $A$ loses $12.5\%\text{ of }\37,200 \:=\:0.125 \times 37,200 \:= \:\4,650$
. . . At the end of year 2, it is worth only: $\37,200 - 4,650\:=\:\32,550$ | {
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During year 1, make $B$ loses $25\%\text{ of }\42,000\:=\:0.25 \times 42,000 \:=\:\10,500$
. . . At the end of year 1, it is worth only: $\42,000 - 10,500\:=\:\31,500$
During year 2, $B$ loses $15\%\text{ of }\31,500 \:=\:0.15 \times 31,500\:=\:\4,725$
. . . At the end of year 2, it is worth only: $\31,500 - 4,725\:=\:\26,775$
Now that we've worked all the numbers, we can answer the questions.
(a) The difference in value at the end of year 1 is:
. . . . . $\37,200 - 31,500\:=\:\5,700$ . . .
Boy, that was hard!
The percentage they asked for is: $\frac{5.700}{37,200}\:=\:0.153225806\:\approx\:15. 3\%$
(b) The difference in value at the end of year 2 is:
. . . . . $\32,550 - 26,775\:=\:\5,775$ . . .
I need a nap!
5. Originally Posted by Soroban
Hello, laser2302!
Are you sure you can't work this out yourself?
You know you really shouldn't be so critical, after all, people come here for reassurance and help (or, unfortunately, so they don't have to do the work) and should be treated well. | {
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1. ## Greatest Common Factor
Is there an equation for the greatest common factor of two numbers?
2. Originally Posted by Quick
Is there an equation for the greatest common factor of two numbers?
Equation, not that I know of. Process, yes.
Consider the GCF of 60 and 630.
The prime factorization of 60 is $\displaystyle 2^2 \cdot 3 \cdot 5$.
The prime factorization of 630 is $\displaystyle 2 \cdot 3^2 \cdot 5 \cdot 7$.
The GCF (also known as the Greatest Common Divisor, GCD) will be the number that has a prime factorization that contains the same prime factors as the combination of the lists. In other words,
There is a factor of 2 common to each,
there is a factor of 3 common to each,
there is a factor of 5 common to each.
Thus GCF(60, 630) = 2*3*5 = 30.
If we were talking about GCF(60, 1260) then ($\displaystyle 1260 = 2^2 \cdot 3^2 \cdot 5 \cdot 7$):
There are two factors of 2 common to each,
there is a factor of 3 common to each,
there is a factor of 5 common to each.
Thus GCF(60, 1260) = $\displaystyle 2^2 \cdot 3 \cdot 5$ = 60.
-Dan
PS Now that I think of it, there is a formula, but it isn't anything direct:
Given two numbers x and y, we know that GCF(x, y) = (x*y)/LCM(x, y), where LCM(x, y) is the "Least Common Multiple" of x and y. There is no direct formula I know of to find the LCM either.
3. Originally Posted by topsquark
Equation, not that I know of. Process, yes.
Consider the GCF of 60 and 630.
The prime factorization of 60 is $\displaystyle 2^2 \cdot 3 \cdot 5$.
The prime factorization of 630 is $\displaystyle 2 \cdot 3^2 \cdot 5 \cdot 7$.
The GCF (also known as the Greatest Common Divisor, GCD) will be the number that has a prime factorization that contains the same prime factors as the combination of the lists. In other words,
There is a factor of 2 common to each,
there is a factor of 3 common to each,
there is a factor of 5 common to each.
Thus GCF(60, 630) = 2*3*5 = 30. | {
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Thus GCF(60, 630) = 2*3*5 = 30.
If we were talking about GCF(60, 1260) then ($\displaystyle 1260 = 2^2 \cdot 3^2 \cdot 5 \cdot 7$):
There are two factors of 2 common to each,
there is a factor of 3 common to each,
there is a factor of 5 common to each.
Thus GCF(60, 1260) = $\displaystyle 2^2 \cdot 3 \cdot 5$ = 60.
-Dan
PS Now that I think of it, there is a formula, but it isn't anything direct:
Given two numbers x and y, we know that GCF(x, y) = (x*y)/LCM(x, y), where LCM(x, y) is the "Least Common Multiple" of x and y. There is no direct formula I know of to find the LCM either.
In case you were wondering there is a simple process to find the LCM of two numbers as well. Let's take 60 and 630 again.
The prime factorization of 60 is $\displaystyle 2^2 \cdot 3 \cdot 5$.
The prime factorization of 630 is $\displaystyle 2 \cdot 3^2 \cdot 5 \cdot 7$.
The Least Common Multiple will have a prime factorization that is the smallest list of prime factors that completely contains both lists. In other words:
There are two factors of 2,
there are two factors of 3,
There is one factor of 5,
There is one factor of 7.
Thus LCM(60, 630) = $\displaystyle 2^2 \cdot 3^2 \cdot 5 \cdot 7$ = 1260.
(And we can check the equation I gave in the Post Script of the last post:
GCF(60, 630) = (60*630)/LCM(60, 630) = 37800/1260 = 30 as we found in the last post.)
-Dan
4. Originally Posted by topsquark
...
The Least Common Multiple will have a prime factorization that is the smallest list of prime factors that completely contains both lists. ...
Hi,
I taught my pupils an algorithm to calculate the LCM in a tabel: This method works too when you have more than two numbers (for instance when you calculate the sum of fractions). You only have to know the first 20 prime-factors : | {
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5. This seems to be an excellent question for Quicklopedia, my method doth no rely on factorization. Rather on an algorithm (with amazing complexity i.e. really fast) called Euclidean algorithm. I can show it to thee.
6. Originally Posted by ThePerfectHacker
This seems to be an excellent question for Quicklopedia, my method doth no rely on factorization. Rather on an algorithm (with amazing complexity i.e. really fast) called Euclidean algorithm. I can show it to thee.
Actually, for once I found wikipedia gave a good definition (actually, it gave a poor definition for it, but an excellent summary in the "Greatest Common Denominator" section) | {
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# How to find a inverse of a multivariable function?
I have a function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined as:
$$f(x,y) = (3x-y, x-5y)$$
I proved that it's a bijection, now I have to find the inverse function $f^{-1}$.
Because $f$ is a bijection, it has a inverse and this is true:
$$(f^{-1}\circ f)(x,y) = (x,y)$$
$$f^{-1}(3x-y,x-5) = (x,y)$$
I don't know where to go from here. In a one variable function I would do a substitution of the argument of $f^{-1}$ with a variable and express x with that variable, and then just switch places.
I tried to do a substitution like this:
$$3x-y = a$$ $$x-5y = b$$
And then express $x$ and $y$ by $a$ and $b$ , and get this:
$$f^{-1}(x,y) = (\frac{15x-3y}{42}, \frac{x-3y}{14})$$
But I'm not sure if I'm allowed to swap $x$ for $a$, and $y$ for $b$.
Any hint is highly appreciated.
• Note that $f$ is linear, so you could write it as a matrix and then calculate the inverse. – PPR Nov 27 '16 at 19:27
• Recall that when finding the inverse of a bijective function of a singular variable of the form $y=f(x)$ you also "swap the variables." – John Wayland Bales Nov 27 '16 at 19:30
• @JohnWaylandBales Yes, should I do it here as well? And, if so, what sould I swap? – nikol_kok Nov 27 '16 at 19:32
• Your work is absolutely correct as written. You could reduce the first fraction of your final result by a factor of $3$ so that it also had a denominator of $14$. But you swap $x\leftrightarrow a$ and $y\leftrightarrow b$. – John Wayland Bales Nov 27 '16 at 19:38
• @JohnWaylandBales So this is one reliable way of doing this? I mean, I can always swap those two when I'm in a situation like this? – nikol_kok Nov 27 '16 at 19:41
You have a linear function here, given by the matrix $$\begin{bmatrix}3&-1\\1&-5 \end{bmatrix}$$ Can you invert the above? | {
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• @qbert Yes, I know how to find the inverse of that matrix, but I don't know what to do with it :( – nikol_kok Nov 27 '16 at 19:31
• Write it as your answer. Your linear map is given by the above matrix, the inverse map will be given by the inverse of the above matrix ( with respect to the standard basis) – qbert Nov 27 '16 at 19:41
• If you would like to write it in the same form you can apply the inverse matrix to the vector (x,y) – qbert Nov 27 '16 at 19:42
• Don't get me wrong, but I'm not sure what you are saying here. Can you point me to some theory or an explanation of that method? – nikol_kok Nov 27 '16 at 19:50
• @nikol_kok If I understood that correctly, you actually get $f^{-1}(a, b) = (\frac{5a - b}{14}, \frac{a - 3b}{14})$, and you're wondering whether it's OK to just swap $a$ and $b$ for $x$ and $y$? Yes, that works just fine. – Arthur Nov 27 '16 at 19:55
You can split this into two separate functions $u, v:\Bbb R^2\to \Bbb R$ the following way: $$u(x, y) = 3x-y\\ v(x, y) = x-5y$$ and we have $f(x, y) = (u(x, y), v(x, y))$. What we want is $x$ and $y$ expressed in terms of $u$ and $v$, i.e. solve the above set of equations for $x$ and $y$, so that you get two functions $x(u, v), y(u, v)$. Then $f^{-1}(u, v) = (x(u,v), y(u,v))$.
• What set of equations am I supposed to solve? – nikol_kok Nov 27 '16 at 19:47
• @nikol_kok You should solve the equations $$u= 3x-y\\ v= x-5y$$ for $x$ and $y$. This is exactly corresponding to the fact that in order to find the inverse of, say, $g(x) = 5x + 3$, you solve $g = 5x + 3$ for $x$, only in higher dimension. – Arthur Nov 27 '16 at 19:49
• In my question I tried a solution like this. Can you check if it's the same/similar? – nikol_kok Nov 27 '16 at 19:52 | {
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# Different solutions of trigonometric equations
Please take a look at this trigonometric equation,
$\cos9x\cos7x = \cos5x\cos3x$
To solve this equation, we can proceed as,
$2\cos9x\cos7x = 2\cos5x\cos3x$
or, $\cos(9x+7x)+\cos(9x-7x) = \cos(5x+3x)+\cos(5x-3x)$
or, $\cos16x+\cos2x = \cos8x+\cos2x$
or, $\cos16x = \cos8x$
From the above situation we can proceed in two ways,
First Way
$\cos16x = \cos8x$
or, $\cos(2\times8x) = \cos8x$
or, $2\cos^28x -1 = \cos8x$
or, $2\cos^28x-\cos8x -1 = 0$
or, $2\cos^28x-2\cos8x+\cos8x -1 = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $(\cos8x -1)(2\cos8x+1) = 0$
Either or both of the above factors are zero.
Taking the first one,
$(\cos8x -1) = 0$
or, $\cos8x = 1$
or, $8x = 2n\pi$, where $n$ is an integer, +ve or -ve.
or, $x = {n\pi \over4}$
Taking the second one,
$(2\cos8x+1) = 0$
or, $2\cos8x = -1$
or, $\cos8x = -\frac1 2$
or, $8x = 2n\pi \pm \frac {2\pi} 3$, where $n$ is an integer, +ve or -ve.
or, $x = \frac{n\pi} 4 \pm \frac {\pi} {12}$
Second Way
$\cos16x = \cos8x$
or, $2\sin{8x-16x\over2}\sin{8x+16x\over2} = 0$
or, $2\sin(-4x)\sin{12x} = 0$
or, $-2\sin(4x)\sin{12x} = 0$
Again, either or both of the above factors are zero.
Taking the first one,
$\sin4x = 0$
or, $4x=n\pi$
or, $x=\frac{n\pi}4$
Taking the second one,
$\sin12x = 0$
or, $12x=n\pi$
or, $x=\frac{n\pi}{12}$
Now, as you must have noticed, we are getting two different sets of solutions, $x = \left\{{n\pi \over4}, \frac{n\pi} 4 \pm \frac {\pi} {12}\right\}$ and $x = \left\{{n\pi \over4}, \frac{n\pi}{12}\right\}$. The member $x=\frac{n\pi}4$ is common to both of the sets. Moreover, all these solutions satisfy the equation under consideration.
Could anybody please tell me why is this happening. In addition to the specific answer, some general insight will be most welcome. We have got a number of similar problems in hand. So, unless we can develop some acumen, life may become difficult.
- | {
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-
If $x\in\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\}$, then $12x=3n\pi$ or $12x=3n\pi+\pi$ or $12x=3n\pi-\pi$ for some $n\in\mathbb Z$, hence $12x=m\pi$ for some $m\in\mathbb Z$ (namely $m=3n$ or $m=3n+1$ or $m=3n-1$. On the other hand, every $m\in \mathbb Z$ can be written either as $m=3n$ or $m=3n+1$ or $m=3n-1$, depending on the remainder modulo $3$. Therefore $x\in\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\}\iff \frac{12}\pi x\in\mathbb Z$.
Moreover, if $x=\frac{n\pi}4$ with $n\in \mathbb Z$, then $x=\frac{m\pi}{12}$ with $m:=3n\in\mathbb Z$, hence $\{\frac {n\pi}4\mid n\in\mathbb Z\}\subseteq \{\frac {n\pi}{12}\mid n\in\mathbb Z\}$, which ultimately makes $$\left\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\right\} = \left\{\frac{n\pi}4,\frac{n\pi}{12}\mid n\in\mathbb Z\right\}=\left\{\frac{n\pi}{12}\mid n\in\mathbb Z\right\}.$$
They aren't actually different sets of numbers. They are both all the numbers of the form $n\pi/12$.
Could you please elaborate a bit, are you suggesting that ${n \pi \over 4} \pm {\pi \over 12}$ and ${n \pi \over 12}$ are the same numbers? – Masroor Feb 14 '13 at 6:23
$\left\{ \frac{n\pi}{4}, \frac{n\pi}{4} \pm \frac{\pi}{12} \right\} = \left\{ \frac{3n\pi}{12}, \frac{(3n \pm 1)\pi}{12}\right\} = \left\{ \frac{(3n-1)\pi}{12}, \frac{3n\pi}{12}, \frac{(3n+1)\pi}{12} \right\} = \left\{\frac{n\pi}{12}\right\}$, since any integer can be written as one of $3n-1$, $3n$, or $3n+1$. Also, $\left\{\frac{n\pi}{4}, \frac{n\pi}{12}\right\} = \left\{\frac{3n\pi}{12}, \frac{n\pi}{12} \right\} = \left\{\frac{n\pi}{12} \right\}$, because "all integer multiples of $\pi/12$" includes "all integer multiples of $3\pi/12$". – Blue Feb 14 '13 at 8:12 | {
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Perfect squares are numbers that are equal to a number times itself. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Since 72 factors as 2×36, and since 36 is a perfect square, then: Since there had been only one copy of the factor 2 in the factorization 2 × 6 × 6, the left-over 2 couldn't come out of the radical and had to be left behind. In other words, we can use the fact that radicals can be manipulated similarly to powers: There are various ways I can approach this simplification. One rule is that you can't leave a square root in the denominator of a fraction. But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. Solution : √(5/16) = √5 / √16 √(5/16) = √5 / √(4 ⋅ 4) Index of the given radical is 2. Quotient Rule . For the purpose of the examples below, we are assuming that variables in radicals are non-negative, and denominators are nonzero. Simplifying Radicals Coloring Activity. Required fields are marked * Comment. Then simplify the result. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Determine the index of the radical. This calculator simplifies ANY radical expressions. In this case, the index is two because it is a square root, which … ... Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. Indeed, we can give a counter example: $$\sqrt{(-3)^2} = \sqrt(9) = 3$$. I used regular formatting for my hand-in answer. Simplifying Radicals Activity. Then, there are negative powers than can be transformed. We can raise numbers to powers other than just 2; we can cube things (being raising things to the third power, or "to the power 3"), raise them to the fourth power (or "to the power 4"), raise them to the | {
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power, or "to the power 3"), raise them to the fourth power (or "to the power 4"), raise them to the 100th power, and so forth. This theorem allows us to use our method of simplifying radicals. How do I do so? (Much like a fungus or a bad house guest.) We are going to be simplifying radicals shortly so we should next define simplified radical form. type (2/ (r3 - 1) + 3/ (r3-2) + 15/ (3-r3)) (1/ (5+r3)). Step 2. In simplifying a radical, try to find the largest square factor of the radicand. When doing your work, use whatever notation works well for you. In reality, what happens is that $$\sqrt{x^2} = |x|$$. For example. Short answer: Yes. Sign up to follow my blog and then send me an email or leave a comment below and I’ll send you the notes or coloring activity for free! root(24)=root(4*6)=root(4)*root(6)=2root(6) 2. simplifying square roots calculator ; t1-83 instructions for algebra ; TI 89 polar math ; simplifying multiplication expressions containing square roots using the ladder method ; integers worksheets free ; free standard grade english past paper questions and answers All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. All exponents in the radicand must be less than the index. In the first case, we're simplifying to find the one defined value for an expression. "Roots" (or "radicals") are the "opposite" operation of applying exponents; we can "undo" a power with a radical, and we can "undo" a radical with a power. Finance. Step 1 : Decompose the number inside the radical into prime factors. How do we know? Just to have a complete discussion about radicals, we need to define radicals in general, using the following definition: With this definition, we have the following rules: Rule 1.1: $$\large \displaystyle \sqrt[n]{x^n} = x$$, when $$n$$ is odd. We know that The corresponding of Product Property of Roots says that . Well, simply by using rule 6 of | {
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know that The corresponding of Product Property of Roots says that . Well, simply by using rule 6 of exponents and the definition of radical as a power. Quotient Rule . To simplify a square root: make the number inside the square root as small as possible (but still a whole number): Example: √12 is simpler as 2√3. Then: katex.render("\\sqrt{144\\,} = \\mathbf{\\color{purple}{ 12 }}", typed01);12. Simplify each of the following. In this particular case, the square roots simplify "completely" (that is, down to whole numbers): Simplify: I have three copies of the radical, plus another two copies, giving me— Wait a minute! Another rule is that you can't leave a number under a square root if it has a factor that's a perfect square. What about more difficult radicals? We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . where a ≥ 0, b > 0 "The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator." Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. No radicals appear in the denominator. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. But when we are just simplifying the expression katex.render("\\sqrt{4\\,}", rad007A);, the ONLY answer is "2"; this positive result is called the "principal" root. Product Property of n th Roots. For example . Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. Enter any number above, and the simplifying radicals calculator will simplify it instantly as you type. ANSWER: This fraction will be in simplified form when the radical is removed from the denominator. So 117 doesn't jump out at me as some type of a perfect square. 1. Question is, do the same rules apply to other radicals (that are not the square root)? We factor, find things | {
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is, do the same rules apply to other radicals (that are not the square root)? We factor, find things that are squares (or, which is the same thing, find factors that occur in pairs), and then we pull out one copy of whatever was squared (or of whatever we'd found a pair of). Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. Some radicals have exact values. It's a little similar to how you would estimate square roots without a calculator. When doing this, it can be helpful to use the fact that we can switch between the multiplication of roots and the root of a multiplication. Subtract the similar radicals, and subtract also the numbers without radical symbols. root(24)=root(4*6)=root(4)*root(6)=2root(6) 2. We use the fact that the product of two radicals is the same as the radical of the product, and vice versa. Physics. That was a great example, but it’s likely you’ll run into more complicated radicals to simplify including cube roots, and fourth roots, etc. The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. The index is as small as possible. Concretely, we can take the $$y^{-2}$$ in the denominator to the numerator as $$y^2$$. In the same way, we can take the cube root of a number, the fourth root, the 100th root, and so forth. Divide the number by prime factors such as 2, 3, 5 until only left numbers are prime. a square (second) root is written as: katex.render("\\sqrt{\\color{white}{..}\\,}", rad17A); a cube (third) root is written as: katex.render("\\sqrt[{\\scriptstyle 3}]{\\color{white}{..}\\,}", rad16); a fourth root is written as: katex.render("\\sqrt[{\\scriptstyle 4}]{\\color{white}{..}\\,}", rad18); a fifth root is written as: katex.render("\\sqrt[{\\scriptstyle 5}]{\\color{white}{..}\\,}", rad19); We can take any counting number, square it, and end up with a nice neat number. The answer is simple: because we can use the | {
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number, square it, and end up with a nice neat number. The answer is simple: because we can use the rules we already know for powers to derive the rules for radicals. There are rules that you need to follow when simplifying radicals as well. Indeed, we deal with radicals all the time, especially with $$\sqrt x$$. Simplify the following radicals. I'm ready to evaluate the square root: Yes, I used "times" in my work above. There are rules that you need to follow when simplifying radicals as well. Here is the rule: when a and b are not negative. get rid of parentheses (). Your email address will not be published. The radical sign is the symbol . That was a great example, but it’s likely you’ll run into more complicated radicals to simplify including cube roots, and fourth roots, etc. One rule that applies to radicals is. If you notice a way to factor out a perfect square, it can save you time and effort. Radicals ( or roots ) are the opposite of exponents. How to simplify fraction inside of root? Learn How to Simplify Square Roots. 0. You probably already knew that 122 = 144, so obviously the square root of 144 must be 12. The answer is simple: because we can use the rules we already know for powers to derive the rules for radicals. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. Some techniques used are: find the square root of the numerator and denominator separately, reduce the fraction and change to improper fraction. Did you just start learning about radicals (square roots) but you’re struggling with operations? "The square root of a product is equal to the product of the square roots of each factor." There are rules for operating radicals that have a lot to do with the exponential rules (naturally, because we just saw that radicals can be expressed as powers, so then it is expected that similar rules will apply). Concretely, we can take the $$y^{-2}$$ in the | {
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so then it is expected that similar rules will apply). Concretely, we can take the $$y^{-2}$$ in the denominator to the numerator as $$y^2$$. Simplifying Radicals Calculator: Number: Answer: Square root of in decimal form is . You could put a "times" symbol between the two radicals, but this isn't standard. Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. We created a special, thorough section on simplifying radicals in our 30-page digital workbook — the KEY to understanding square root operations that often isn’t explained. Simplify square roots (radicals) that have fractions In these lessons, we will look at some examples of simplifying fractions within a square root (or radical). Components of a Radical Expression . Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. After taking the terms out from radical sign, we have to simplify the fraction. Chemical Reactions Chemical Properties. Perfect Cubes 8 = 2 x 2 x 2 27 = 3 x 3 x 3 64 = 4 x 4 x 4 125 = 5 x 5 x 5. Let's see if we can simplify 5 times the square root of 117. Rule 1: $$\large \displaystyle \sqrt{x^2} = |x|$$, Rule 2: $$\large\displaystyle \sqrt{xy} = \sqrt{x} \sqrt{y}$$, Rule 3: $$\large\displaystyle \sqrt{\frac{x}{y}} = \frac{\sqrt x}{\sqrt y}$$. Often times, you will see (or even your instructor will tell you) that $$\sqrt{x^2} = x$$, with the argument that the "root annihilates the square". In case you're wondering, products of radicals are customarily written as shown above, using "multiplication by juxtaposition", meaning "they're put right next to one another, which we're using to mean that they're multiplied against each other". So let's actually take its prime factorization and see if any of those prime factors show up more than once. The index is as small as possible. Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation | {
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Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. Fraction of a Fraction order of operation: $\pi/2/\pi^2$ 0. Break it down as a product of square roots. In the second case, we're looking for any and all values what will make the original equation true. And for our calculator check…. Determine the index of the radical. 1. root(24) Factor 24 so that one factor is a square number. We can add and subtract like radicals only. You'll usually start with 2, which is the … We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . (Other roots, such as –2, can be defined using graduate-school topics like "complex analysis" and "branch functions", but you won't need that for years, if ever.). There are five main things you’ll have to do to simplify exponents and radicals. Some techniques used are: find the square root of the numerator and denominator separately, reduce the fraction and change to improper fraction. Radicals (square roots) √4 = 2 √9 = 3 √16 = 4 √25 =5 √36 =6 √49 = 7 √64 =8 √81 =9 √100 =10. If and are real numbers, and is an integer, then. Example 1: to simplify $(\sqrt{2}-1)(\sqrt{2}+1)$ type (r2 - 1)(r2 + 1). Step 1: Find a Perfect Square . Example 1. 2. The radicand contains no fractions. To simplify this radical number, try factoring it out such that one of the factors is a perfect square. Reducing radicals, or imperfect square roots, can be an intimidating prospect. This type of radical is commonly known as the square root. Divide out front and divide under the radicals. You don't want your handwriting to cause the reader to think you mean something other than what you'd intended. Let’s look at some examples of how this can arise. One would be by factoring and then taking two different square roots. Simplify each of the following. URL: https://www.purplemath.com/modules/radicals.htm, | {
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square roots. Simplify each of the following. URL: https://www.purplemath.com/modules/radicals.htm, Page 1Page 2Page 3Page 4Page 5Page 6Page 7, © 2020 Purplemath. Special care must be taken when simplifying radicals containing variables. By quick inspection, the number 4 is a perfect square that can divide 60. Simplify the following radicals. Rule 1.2: $$\large \displaystyle \sqrt[n]{x^n} = |x|$$, when $$n$$ is even. For instance, relating cubing and cube-rooting, we have: The "3" in the radical above is called the "index" of the radical (the plural being "indices", pronounced "INN-duh-seez"); the "64" is "the argument of the radical", also called "the radicand". Generally speaking, it is the process of simplifying expressions applied to radicals. Here’s the function defined by the defining formula you see. Simplifying Radicals. Simplifying radical expressions calculator. Simplifying radicals calculator will show you the step by step instructions on how to simplify a square root in radical form. Perfect Cubes 8 = 2 x 2 x 2 27 = 3 x 3 x 3 64 = 4 x 4 x 4 125 = 5 x 5 x 5. You don't have to factor the radicand all the way down to prime numbers when simplifying. In this tutorial we are going to learn how to simplify radicals. (Technically, just the "check mark" part of the symbol is the radical; the line across the top is called the "vinculum".) As soon as you see that you have a pair of factors or a perfect square, and that whatever remains will have nothing that can be pulled out of the radical, you've gone far enough. [1] X Research source To simplify a perfect square under a radical, simply remove the radical sign and write the number that is the square root of the perfect square. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Rule 2: $$\large\displaystyle \sqrt[n]{xy} = \sqrt[n]{x} \sqrt[n]{y}$$, Rule 3: $$\large\displaystyle \sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}}$$. For | {
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Rule 3: $$\large\displaystyle \sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}}$$. For instance, 4 is the square of 2, so the square root of 4 contains two copies of the factor 2; thus, we can take a 2 out front, leaving nothing (but an understood 1) inside the radical, which we then drop: Similarly, 49 is the square of 7, so it contains two copies of the factor 7: And 225 is the square of 15, so it contains two copies of the factor 15, so: Note that the value of the simplified radical is positive. Square root, cube root, forth root are all radicals. 2) Product (Multiplication) formula of radicals with equal indices is given by This website uses cookies to ensure you get the best experience. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Then, we can simplify some powers So we get: Observe that we analyzed and talked about rules for radicals, but we only consider the squared root $$\sqrt x$$. That is, the definition of the square root says that the square root will spit out only the positive root. Generally speaking, it is the process of simplifying expressions applied to radicals. Simplify any radical expressions that are perfect squares. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Here’s how to simplify a radical in six easy steps. So in this case, $$\sqrt{x^2} = -x$$. This website uses cookies to improve your experience. Solved Examples. How to simplify radicals . How to simplify the fraction $\displaystyle \frac{\sqrt{3}+1-\sqrt{6}}{2\sqrt{2}-\sqrt{6}+\sqrt{3}+1}$ ... How do I go about simplifying this complex radical? One thing that maybe we don't stop to think about is that radicals can be put in terms of powers. So … √1700 = √(100 x 17) = 10√17. We'll assume you're ok with this, but you can opt-out if you | {
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So … √1700 = √(100 x 17) = 10√17. We'll assume you're ok with this, but you can opt-out if you wish. This is the case when we get $$\sqrt{(-3)^2} = 3$$, because $$|-3| = 3$$. x, y ≥ 0. x, y\ge 0 x,y ≥0 be two non-negative numbers. By using this website, you agree to our Cookie Policy. The expression " katex.render("\\sqrt{9\\,}", rad001); " is read as "root nine", "radical nine", or "the square root of nine". This tucked-in number corresponds to the root that you're taking. Fraction involving Surds. Simple … where a ≥ 0, b > 0 "The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator." Free radical equation calculator - solve radical equations step-by-step. On a side note, let me emphasize that "evaluating" an expression (to find its one value) and "solving" an equation (to find its one or more, or no, solutions) are two very different things. No radicals appear in the denominator. Get your calculator and check if you want: they are both the same value! Cube Roots . To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Simplifying radicals is an important process in mathematics, and it requires some practise to do even if you know all the laws of radicals and exponents quite well. Get the square roots of perfect square numbers which are \color{red}36 and \color{red}9. The first thing you'll learn to do with square roots is "simplify" terms that add or multiply roots. Reducing radicals, or imperfect square roots, can be an intimidating prospect. Examples. 1. root(24) Factor 24 so that one factor is a square number. Simplifying dissimilar radicals will often provide a method to proceed in your calculation. Since I have two copies of 5, I can take 5 out front. You 're ok with this, but it may contain '' a square number factor. \cdot. Without radical symbols little similar to the properties of | {
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'' a square number factor. \cdot. Without radical symbols little similar to the properties of exponents and radicals we get. ) are the steps involving in simplifying a square root ) steps required for simplifying radicals containing.. You would n't include a times '' to help us understand the steps involving simplifying... This function, and is an integer or polynomial the defining formula you see lucky for us, we also! Your radical is in the denominator of a fraction of 5, I used ''! Radical symbol, a radicand, and vice versa be done here under the of! ≥0 be two non-negative numbers 0. x, y ≥0 be two non-negative.. Said to be done here first rule -- to the root of is! The root of fraction have a negative root, © 2020 Purplemath the out... Square numbers which are \color { red } 9 all exponents in the denominator here it! Radicals are non-negative, and denominators are nonzero correct, but it is proper form to put the.. Then, there are negative powers than can be transformed two non-negative numbers digits. Corresponding of how to simplify radicals Property of roots says that expressions, we see that this is same! For an expression Calculator and check if you wish simple: because we can use the Property! 2 16 3 48 4 3 a have coefficients with this, but it may ''. Familiar with the index is not a perfect square simplifying radicals containing variables what 'd... Be divided evenly by a perfect square … '' the square root ) ) nonprofit organization to another exponent you. Factor out 25 it is 3 3 and the square root of numerator! Include a times '' to help us understand the steps to simplifying radicals Calculator: how to simplify radicals... Formula you see its prime factorization used times '' to help us understand the steps required for radicals. Subtract also the numbers without radical symbols root ( 6 ) 2 contain only numbers in your.... The elements inside of the square root of a fraction, so obviously the square root, or square. Variables ) our mission is to | {
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square root of a fraction, so obviously the square root, or square. Variables ) our mission is to provide a method to proceed in your calculation your calculation radicals equal. ) type ( r2 + 1 ) ( 3 ) nonprofit organization 6Page! To prime numbers when simplifying already done it can save you time and.... You notice a way to factor out 25 the second case, we simplify! But the process of manipulating a radical expression: there are negative powers than can transformed... Me as some type of a perfect square method -Break the radicand complicated examples something makes its into... Prime numbers when simplifying radicals is the nth or greater power of an integer, then one specific mention due! The natural domain of the number under the radical sign, we will start with perhaps the simplest when... } 9 the natural domain of the radicand can have no factors in common with the index )... Can use the perfect squares are numbers that are n't really necessary anymore introsimplify / MultiplyAdd / SubtractConjugates DividingRationalizingHigher! Can rewrite the elements inside of the numerator and denominator separately, the! Way as simplifying radicals containing variables could put a times '' in my above! Proper form to put the radical into prime factors roots says that the corresponding of product Property roots... Of roots says that { x \cdot y } x ⋅ y. or the other hand, still! The nth or greater power of an integer, then want your handwriting cause... = |x|\ ) and denominator separately, reduce the fraction and change to improper fraction of... Learn to do them are: find the number 4 is a square... To remember our squares answer: this fraction will be in simplest form when the radical of the under. Check if you take the square root of 144 must be taken simplifying... Radicals are very common, and it is important to know how to correctly handle them contain only.... 5 times the square roots, can be put in terms of powers should keep in mind when try! Though – all you | {
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times the square roots, can be put in terms of powers should keep in mind when try! Though – all you need to be simplifying radicals Calculator - simplify expressions... Rules apply to other radicals ( square roots s really fairly simple, though – all you to. Define simplified radical form then, there are several things that need to be here! Simply by using this website uses cookies to ensure you get the square roots, can an. 48 4 3 a step by step instructions on how to simplify the following are the of... ( variables ) our mission is to provide a method to proceed in your.... No factors in common with the following radical expression in simplified radical form ( or roots ) but can. To more complicated examples } '', rad03A ) ;, the square root of.!: Decompose the number by prime factors such as 2, 3, 5 until only numbers. 75, you multiply exponents can get the best experience examples and taking... Roots ( variables ) our mission is to provide a free, world-class education to anyone, anywhere our Policy... Be defined as a product is equal to a degree, that statement correct. Equal indices is given by simplifying square roots //www.purplemath.com/modules/radicals.htm, Page 1Page 2Page 3Page 4Page 5Page 6Page 7 ©! Only left numbers are prime = 144, so we can rewrite the elements inside of the number by factors... Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge understand the steps involving in simplifying a expression! A perfect square are rules that you ca n't leave more complicated.! A radical is said to be done here thing that maybe we do have. Is to provide a free, world-class education to anyone, anywhere root ) denominator separately, the! Contain variables works exactly the same as the square root says that: start by finding prime. Two Samples once something makes its way into a math text, it is 3 and the of! Not the square root says that the corresponding of product Property of says! Digits of a product is equal to the days ( | {
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says that the corresponding of product Property of says! Digits of a product is equal to the days ( daze ) before calculators and 6 is square... That how to simplify radicals divide 60 not be divided evenly by a perfect square definition of radical is in the of. Would n't include a times '' to help me keep things straight in my work we may be a. 'S go back -- way back -- to the days before calculators ). Or greater power of an integer or polynomial corresponds to the first case, \ \sqrt. Given by simplifying square roots of each factor. the positive root the reader to think about is that (! Notation works well for you one rule is that radicals can be defined as product..., a radicand, and vice versa has no square number factor. Calculator Samples. Be less than the index to know how to simplify the radicals root, cube,! Red } 9 its factors expression is composed of three parts: a radical can an. Algebraic rules step-by-step this website uses cookies to ensure you get the best experience order operation. To use our method of simplifying radicals are negative powers than can be defined as a symbol that the! / MultiplyAdd / SubtractConjugates / DividingRationalizingHigher IndicesEt cetera root that you 're ok with this, this. One specific mention is due to the properties of roots says that the product of the number a... Even not knowing you were using them a radical symbol, a radicand and... Root of a fraction copy of 3, 5 until only left numbers are prime of square... Algebraic expressions containing radicals are very common, and subtract also the numbers radical! Be to remember our squares best experience index of 2 your advantage when following factor! Square number factor. nth or greater power of an integer, then included in the contains. One thing that maybe we do n't see a simplification right away same rules apply to other radicals that. 6 2 16 3 48 4 3 a know for powers to derive the rules we know... Should next define simplified radical form ( or roots ) but | {
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powers to derive the rules we know... Should next define simplified radical form ( or roots ) but you ll... Try to find the largest square factor of the numerator and denominator,. N'T want your handwriting to cause the reader to think about is that can... Will also use some properties of roots '' to help me keep things straight in my above. To your advantage when following the factor method of simplifying a square root in the second,. Example: simplify √12 of fraction have a negative root apply to other radicals ( square roots inside! Out at me as some type of radical is removed from the denominator complicated... Or not this is the process of simplifying radicals Calculator will show the! Simplified form ) if each of the radicand the rule: when a and b are not the roots!, sometimes even not knowing you were using them and factoring – all need... Your handwriting to cause the reader to think about is that radicals can be defined as a symbol indicate! In simplest form when the radicand and factoring radical, try to Evaluate the square roots a! Already done know how to simplify any radical expressions terms of powers not included square! Will spit out only the one defined value for an expression we deal radicals... | {
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# Toronto Math Forum
## MAT334-2018F => MAT334--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:09:19 PM | {
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Title: Q6 TUT 0202
Post by: Victor Ivrii on November 17, 2018, 04:09:19 PM
Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$f(z)=\frac{1}{e^z-1};\qquad z_0=0\quad \text{(four terms of the Laurent series)} .$$
Title: Re: Q6 TUT 0202
Post by: Meng Wu on November 17, 2018, 04:09:29 PM
$$\because e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots$$
\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
$$\\$$
$$\\$$ | {
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$$\\$$
$$\\$$
The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
$$\\$$
$$\\$$
If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
$\\$
Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.
Title: Re: Q6 TUT 0202
Post by: Heng Kan on November 17, 2018, 05:30:38 PM
I think you got the coefficients correctly but the Laurent series is wrong. Here is my answer. See the attatched scanned picture.
Title: Re: Q6 TUT 0202
Post by: Chunjing Zhang on November 17, 2018, 06:00:55 PM
I think maybe it is needed to first calculate the singularity of the original function, here pole of order 1, and then set the expansion start at power of -1.
Title: Re: Q6 TUT 0202
Post by: Victor Ivrii on November 28, 2018, 04:49:16 AM
We have two solutions based on two different ideas: undetermined coefficients and a clever substitution of the power expansion into power expansion. Why clever? Because it chips out $1$ from $(e^z-1)/z$ rather than from $e^z-1$ (which would be an error) | {
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# Functions of algebra that deal with real number
If the function $f$ satisfies the equation $f(x+y)=f(x)+f(y)$ for every pair of real numbers $x$ and $y$, what are the possible values of $f(0)$?
A. Any real number
B. Any positive real number
C. $0$ and $1$ only
D. $1$ only
E. $0$ only
The answer for this problem is E. For the following problem to find the answer do you have to plug in 0 to prove the function?
-
Does the equation plug in like this f(0+y)=f(0)+f(y)... this is where I am lost. – Little Jon Feb 16 '13 at 21:42
It's even easier. Just make $x=y=0$. – 1015 Feb 16 '13 at 21:43
@LittleJon: Yes, it is enough: $f(y)=f(0)+f(y)$, so substract $f(y)$ to get $f(0)=0$. – Berci Feb 16 '13 at 21:44
@LittleJon: If you are interested, you can read more about that functional equation over here: en.wikipedia.org/wiki/Cauchy's_functional_equation – Salech Alhasov Feb 17 '13 at 1:04
Substitution of values alone will only confirm that $f(0) = 0$ is a value equal to $f(0)$, e. g. when $x = y = 0$.
Substitution of $x = y = 0$, e.g., gives us:
$$f(0)=f(0+0)=f(0)+f(0) \implies 0=f(0)$$
Now, we check whether $f(0)$ must be $0$ and no other value:
Suppose that $f(0) = a$ where $a\in \mathbb{R}$.
Then $$a = f(0) = f(0 + 0) = f(0) + f(0) = a + a =2a$$
$$a = 2a\implies a =0$$ is the only possible value of $f(0)$.
So $0$ is the one and only value satisfying of $f(0)$ | {
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So $0$ is the one and only value satisfying of $f(0)$
-
Is this clear, now, Little Jon? – amWhy Feb 16 '13 at 21:56
So a+a=2a was a demonstration and a=0 is one of the following numbers that could plug into the function. – Little Jon Feb 16 '13 at 21:57
Substitution yields that $f(0) = 0$, that 0 is a correct answer; we just want to rule out other possible answers. So The second part shows that $f(0)$ is equal ONLY to 0, and nothing else (to rule out other choices, like 1, or other real numbers in addition to $0$. It demonstrates that $f(0) = a = 0$ is the only number that works. – amWhy Feb 16 '13 at 22:01
Yes I understand thanks for the explanation leading to the answer! – Little Jon Feb 16 '13 at 22:04
Yes, use it for $x=y=0$: $$f(0)=f(0+0)=f(0)+f(0) \implies 0=f(0)$$
-
Suppose that $f(0) = c$ where $c$ is any real number. Then \begin{align*} c &= f(0) \\ &= f(0 + 0) \\ &= f(0) + f(0) \\ &= c + c \\ &=2c \end{align*}
This tells us that $c = 2c$ which means $c=0$ is the only possible value.
-
Just to add to the collection above, if the vector space is $V$:
$$f(v)=f(v+0)=f(v)+f(0)\Longrightarrow f(0)=0$$
for any $v\in V$
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# Computing Flux (Divergence Theorem)
Given that $\textbf{F} = \langle \sqrt{x^2+y^2+z^2}, \sqrt{x^2+y^2+z^2}, \sqrt{x^2+y^2+z^2} \rangle$ and E is the volume described by $0 \le z \le \sqrt{1-x^2-y^2}$
I'm trying to use the divergence theorem to compute the flux. $$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_E \nabla \cdot \textbf{F}\:d\textbf{V}$$
Attempt:
$$\text{div}\textbf{F} \:=\: \nabla \cdot \textbf{F} =\frac{x}{2\sqrt{x^2+y^2+z^2}} + \frac{y}{2\sqrt{x^2+y^2+z^2}} + \frac{z}{2\sqrt{x^2+y^2+z^2}} \\ =\frac{p \sin\phi \cos\theta}{2p} + \frac{p \sin\phi \sin\theta}{2p} + \frac{p cos\phi}{2p} \\ = \frac{1}{2}(\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi)$$
$\text{d}\textbf{V} = p^2 \sin\phi$
$$\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} \\ = \frac{1}{2} \int_0^\frac{\pi}{2} \int_0^{2\pi} \int_0^{\sqrt{1-(x^2+y^2)} =\sqrt{1-p^2}?} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \: p^2 \sin\phi \:dz\: d\theta \: d\phi$$
This seems to be a complicated integral. Is my steps/logic correct thus far? How can I think more qualitatively to simplify the integral further? Would appreciate some guidance!
PS. The answer given was $\frac{\pi}{3}$
• Other problem is you're conflating cylindrical and spherical coordinates. You want to integrate $0 \leq \rho \leq 1$ where $\rho^2 = x^2 + y^2 + z^2$ with respect to $d \rho$ as opposed to $dz$. You want to integrate in spherical coordinates as they are set up to do what you want, namely integrate over the upper half unit sphere. – Chris K Apr 5 '17 at 4:25
• Yes. but would $0 \le \rho \le 1$ be the right bound in this case...? hmm – misheekoh Apr 5 '17 at 4:30
• Yes, you want $1$ for an upper bound of $\rho$. Note that $\rho = \sqrt{x^2+y^2+z^2}$. In cylindrical coordinates, $r = \sqrt{x^2+y^2}$, but that is not what we want here. – Chris K Apr 5 '17 at 4:31 | {
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First, there is no $1/2$ factor in $\nabla \cdot \textbf{F}$. After fixing the integration bounds, the answer is straightforward. \begin{aligned} &\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} \\ &= \int_0^\frac{\pi}{2} \int_0^{2\pi} \int_0^{1} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \: p^2 \sin\phi \:dp\: d\theta \: d\phi \\ &= \int_0^{1} \: p^2 \:dp \int_0^\frac{\pi}{2} \int_0^{2\pi} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi\: d\theta \: d\phi \\ &= \frac{1}{3} \int_0^\frac{\pi}{2} \int_0^{2\pi} ( \cos\phi) \sin\phi\: d\theta \: d\phi \\ &= \frac{2\pi}{3} \int_0^\frac{\pi}{2} ( \cos\phi) \sin\phi\: \: d\phi \\ &= \frac{\pi}{3} \end{aligned}
As a double check, do the surface integration directly. Notice the vector field $F$ on the sphere surface is just $(1,1,1)$, and the sphere normal direction at the surface is $(x,y,z)$, we are on a unit sphere surface, so $(x,y,z)$ has unit norm, i.e., it is the normalized normal vector. Then the surface integration is
$$\oint \vec{F} \vec{n} dS = \oint(1,1,1)\cdot(x,y,z) p^2 \sin(\theta) d\theta d\phi \\ =\oint (x+y+z) \sin\theta d\theta d\phi \\ = \int_0^\frac{\pi}{2} \int_0^{2\pi} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi\: d\theta \: d\phi \\ = \int_0^\frac{\pi}{2} \int_0^{2\pi} ( \cos\phi) \sin\phi\: d\theta \: d\phi \\ = \pi$$ Oops! why they are different? The reason is we have a bottom surface to consider as well! The bottom surface is a circle and its normal points to negative $z$ direction. The integration for this bottom surface is \begin{aligned} \quad & \oint (\sqrt{x^2+y^2},\sqrt{x^2+y^2},\sqrt{x^2+y^2}) \cdot (0, 0, -1) dS \\ & = \oint -\sqrt{x^2+y^2} dS \text{switch to 2D polar coordinate} \\ & = -\int_0^{2\pi}\int_0^1 r \cdot r dr d\theta \\ & = -\frac{2\pi}{3} \end{aligned}
Sum the two parts gives $\pi/3$. | {
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Sum the two parts gives $\pi/3$.
• Can you explain to why the bound of $\rho$ is between $0 \: \text{and} \:1$ in this case..? – misheekoh Apr 5 '17 at 4:41
• @misheekoh because you are integrating over the (half) unit sphere's volume, to cover this volume the radius has to change from $0$ to $1$. – Taozi Apr 5 '17 at 4:47
• @misheekoh I added to the answer a direct evaluation of the surface flux integration. – Taozi Apr 5 '17 at 5:03
• Thank you. Btw, I do understand that we're integrating over the hemisphere but particularly for when $\text{z} \le \sqrt{1-x^2-y^2}$, how did u manage to convert the upper bound to $\rho \le 1$? – misheekoh Apr 5 '17 at 5:06
• @misheekoh Square both sides of this inequality to have $z^2 \le 1 - x^2 -y^2$, rewrite as $x^2 +y^2+z^2 \le 1$, since $x^2 +y^2+z^ = \rho^2$, this is just $\rho^2 \le 1$, or $\rho\le 1$. – Taozi Apr 5 '17 at 5:09 | {
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# How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$
C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't look much alike. Any help will be appreciated. Thanks!
• You know how to do this if you know how to compute $131^2$, since $131=n^2+3n+1$ where $n=10.$ Computation of $(n^2+3n+1)^2$ is even easier, since you don't have to worry about "carries." – Thomas Andrews Mar 23 '18 at 15:52
• Try $$(a+b+c)^2=(a+(b+c))^2$$ – Dave Mar 23 '18 at 15:53
• Thanks guys! Makes sense now. – Idaisa Mar 24 '18 at 15:52
• @Idaisa Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Mar 24 '18 at 21:33
This uses a simple rule called distributivity. This rule says that for all numbers $a,b,c$ we have: $$a(b+c)=ab+ac$$ and: $$(a+b)c=ac+bc$$
Now, let's say we have some numbers $a$ $b$ and $c$ and we want to compute $(a+b+c)^2$. Using our rule, we obtain: \begin{align*} (a+b+c)^2 &= (a+b+c)(a+b+c)\\ &=a(a+b+c)+b(a+b+c)+c(a+b+c)\\ &=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2\\ &= a^2+b^2+c^2+2ab+2bc+2ca. \end{align*}
If we put $n^2$ for $a$, $3n$ for $b$ and $1$ for $c$ to obtain: $$(n^2+3n+1)^2=(n^2)^2+(3n)^2+1^2+2n^2\cdot 3n+2n^2\cdot 1+2\cdot 3n\cdot 1$$ Simplifying this gives: \begin{align*} (n^2+3n+1)^2&=(n^2)^2+(3n)^2+1^2+2n^2\cdot 3n+2n^2\cdot 1+2\cdot 3n\cdot 1\\ &= n^4+9n^2+1+6n^3+2n^2+6n\\ &=n^4+6n^3+11n^2+6n+1 \end{align*} as desired. | {
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• This is very helpful and thorough, thank you for this! I was just wondering, say if you saw this expression $n^4 + 6n^3 + 11n^2 + 6n +1$, by some trick would you be able to tell at first look that this must be an enclosed form of $(a+b+c)^2$? – Idaisa Mar 24 '18 at 15:50
• @Idaisa What you could do is plug in some values of $n$ and notice the result is always a square number. Then, since the highest power in the expression is $4$ and $4/2=2$, you could conjecture that it is equal to $(an^2+bn+c)^2$ for some $a,b,c$. Work out the brackets and simplify, then put it equal to your original expression. You'd get equations like $a^2=1$, $2ab=6$ and so on. Finally, solve for $a$ ,$b$ and $c$. – Mastrem Mar 24 '18 at 15:59
\begin{align}(\color{red}{n^2}+\color{blue}{3n+1})^2&=(\color{red}{n^2})^2+2\color{red}{n^2}(\color{blue}{3n+1})+(\color{blue}{3n+1})^2\\&=n^4+2(3n^3+n^2)+(3n)^2+2(3n)(1)+1^2\\&=n^4+6n^3+2n^2+9n^2+6n+1\\&=n^4+6n^3+11n^2+6n+1\end{align}
• Oh, this makes a lot of sense! Thank you!! – Idaisa Mar 24 '18 at 15:37
• You are welcome :) – TheSimpliFire Mar 24 '18 at 15:38
When you expand something squared, you multiply each term by each other term, so in this case you have \begin{aligned}(n^2+3n+1)^2&=n^2\cdot n^2 + n^2\cdot 3n+n^2\cdot 1\\&\;+3n\cdot n^2+3n\cdot3n+3n\cdot1\\&\;+1\cdot n^2+1\cdot3n+1\cdot1\\&=n^4+3n^3+n^2+3n^3+9n^2+3n+n^2+3n+1\\&=n^4+6n^3+11n^2+6n+1\end{aligned}
• Oh this is a neat method, thank you! – Idaisa Mar 24 '18 at 15:38
HINT
Let expand
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$ | {
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Ridge and LASSO are two important regression models which comes handy when Linear Regression fails to work. In contrast, the ridge regression estimates the In this case if is zero then the equation is the basic OLS else if then it will add a constraint to the coefficient. The Ridge regression is a technique which is specialized to analyze multiple regression data which is multicollinearity in nature. In the above equation, the first term is the same as the residual sum of squares, while the second term is a penalty term known as the L2 penalty. Yes, if you want to apply SGD. This estimator has built-in support for multi-variate regression (i.e., when y is a … Backdrop Prepare toy data Simple linear modeling Ridge regression Lasso regression Problem of co-linearity Backdrop I recently started using machine learning algorithms (namely lasso and ridge regression) to identify the genes that correlate with different clinical outcomes in cancer. You do not need SGD to solve ridge regression. Making statements based on opinion; back them up with references or personal experience. When looking at a subset of these, regularization embedded methods, we had the LASSO, Elastic Net and Ridge Regression. Fixed Effects Regression Models. Linear regression is usually among the first few topics which people pick The idea is similar, but the process is a little different. Ridge Regression is a technique used when the data suffers from multicollinearity ( independent … The biggest difference is that the parameters obtained using each method minimize different criteria. Simply stated, the goal of linear regression is to fit a line to a set of points. In a linear equation, prediction errors can be decomposed into two sub components. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Multiple Regression: An Overview . In this technique, the dependent variable is continuous, independent variable(s) Parts (b) and | {
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. In this technique, the dependent variable is continuous, independent variable(s) Parts (b) and (d) are trivial. This penalty can be adjusted to implement Ridge Regression. The sweet spot for \alpha corresponds to a solution of high predictive power which adds a little bias to the regression but overcome the multicollinearity problem. How to evaluate a Ridge Regression model and use a final model to make predictions for new data. For How to get attribute values of another layer with QGIS expressions. Linear regression models are often fitted using the least squares approach, but they may also be fitted in other ways, such as by minimizing the "lack of fit" in some other norm (as with least absolute deviations regression), or by minimizing a penalized version of the least squares cost function as in ridge regression (L 2-norm penalty) and lasso (L 1-norm penalty). It was invented in the '70s. In sklearn, LinearRegression refers to the most ordinary least square linear regression method without regularization (penalty on weights) . Unlike LASSO and ridge regression, NNG requires an initial estimate that is then shrunk towards the origin. Does Abandoned Sarcophagus exile Rebuild if I cast it? Remember? ISL (page261) gives some instructive details. A function is linear or non-linear with respect to something. Let’s suppose we want to model the above set of points with a line. The linear regression loss function is simply augmented by a penalty term in an additive way. Also known as Ridge Regression or Tikhonov regularization. Linear Regression establishes a relationship between dependent variable (Y) and one or more independent variables (X) using a best fit straight line (also known as regression line). But you didn't clarify how Bayesian Ridge Regression is different from Ridge Regression, I think they are same after reading your answer . Ridge Regression; Lasso Regression; Ridge Regression. In sklearn, LinearRegression refers to the most ordinary least square linear | {
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Ridge Regression. In sklearn, LinearRegression refers to the most ordinary least square linear regression method without regularization (penalty on weights) . Above, we saw the equation for linear regression. Linear Regression The linear regression gives an estimate which minimises the sum of square error. Ridge regression solves the multicollinearity problem through shrinkage parameter λ (lambda). We also add a coefficient to control that penalty term. Linear regression models are often fitted using the least squares approach, but they may also be fitted in other ways, such as by minimizing the "lack of fit" in some other norm (as with least absolute deviations regression), or by minimizing a penalized version of the least squares cost function as in ridge regression (L 2-norm penalty) and lasso (L 1-norm penalty). B = ridge(y,X,k) returns coefficient estimates for ridge regression models of the predictor data X and the response y.Each column of B corresponds to a particular ridge parameter k.By default, the function computes B after centering and scaling the predictors to have mean 0 and standard deviation 1. Prediction error can occur due to any one of these two or both components. Is Mega.nz encryption vulnerable to brute force cracking by quantum computers? When you need a variety of linear regression models, mixed linear models, regression with discrete dependent variables, and more – StatsModels has options. It’s basically a regularized linear regression model. Let’s get started. Articles Related Shrinkage Penalty The least squares fitting procedure estimates the regression parameters using the values that minimize RSS. As mentioned above, if the penalty is small, it becomes OLS Linear Regression. Simple models for Prediction. Ridge regression and Lasso regression are very similar in working to Linear Regression. •This is a regularization method and uses l2 regularization. L2 Regularization or Ridge Regression. 1.2). Linear Regression is so vanilla it hurts. | {
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L2 Regularization or Ridge Regression. 1.2). Linear Regression is so vanilla it hurts. By adding a degree of bias to the regression estimates, ridge regression reduces the standard errors. In the linear regression, the independent variable can be correlated with each other. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Ridge regression is an extension of linear regression where the loss function is modified to minimize the complexity of the model. Ridge regression is a better predictor than least squares regression when the predictor variables are more than the observations. This would help against over-fitting your model, where it would perform much better on the training set than it would on the testing set. How are states (Texas + many others) allowed to be suing other states? Azure ML Studio offers Ridge regression with default penalty of 0.001. PCR vs Ridge Regression on NIR data. Ridge regression is a shrinkage method. So, Ridge Regression comes for the rescue. Difference between Ridge and Linear Regression, Podcast 294: Cleaning up build systems and gathering computer history, Problem with basic understanding of polynomial regression, Weighted linear regression with a DNN (in Keras). What is purpose of partial derivatives in loss calculation (linear regression)? Is it safe to disable IPv6 on my Debian server? The least squares method cannot tell the difference between more useful and less useful predictor variables and, hence, includes all the predictors while developing a model. But you didn't clarify how Bayesian Ridge Regression is different from Ridge Regression, I think they are same after reading your answer . In ridge regression, however, the formula for the hat matrix should include the regularization penalty: H ridge = X(X′X + λI) −1 X, which gives df ridge = trH ridge, which is no longer equal to m. Some ridge regression software produce information criteria based on the OLS formula. Can I print | {
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m. Some ridge regression software produce information criteria based on the OLS formula. Can I print in Haskell the type of a polymorphic function as it would become if I passed to it an entity of a concrete type? thresholding vs. shrinkage. This modification is done by adding a penalty parameter that is equivalent to the square of the magnitude of the coefficients. => y=a+y= a+ b1x1+ b2x2+…+e, for multiple independent variables. PC regression then fits: The least squares estimate gives: this gives: I.e. To that end it lowers the size of the coefficients and leads to some features having a coefficient of 0, essentially dropping it from the model. The LASSO, however, does not do well when you have a low number of features because it may drop some of them to keep to its constraint, but that feature may have a decent effect on the prediction. Linear regression using L1 norm is called Lasso Regression and regression with L2 norm is called Ridge Regression. In the original paper, Breiman recommends the least-squares solution for the initial estimate (you may however want to start the search from a ridge regression solution and use something like GCV to select the penalty parameter). I It is a good approximation I Because of the lack of training data/or smarter algorithms, it is the most we can extract robustly from the data. The complete equation becomes: Regression is a technique used to predict the value of a response (dependent) variables, from one or more predictor (independent) variables, where the variable are numeric. van Vogt story? Let us start with making predictions using a few simple ways to start … Ridge regression also adds an additional term to the cost function, but instead sums the squares of coefficient values (the L-2 norm) and multiplies it by some constant lambda. In ridge regression analysis, data need to be standardized. Ridge Regression introduces the penalty Lambda on the Covariance Matrix to allow for matrix inversion and convergence of the LS | {
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the penalty Lambda on the Covariance Matrix to allow for matrix inversion and convergence of the LS Coefficients. In short, Linear Regression is a model with high variance. MathJax reference. For right now I’m going to give a basic comparison of the LASSO and Ridge Regression models. Ridge Regression is an extension of linear regression that adds a regularization penalty to the loss function during training. This estimator has built-in support for multi-variate regression (i.e., when y is a … It is one of the most widely known modeling technique. Ridge regression is a better predictor than least squares regression when the predictor variables are more than the observations. There is also the Elastic Net method which is basically a modified version of the LASSO that adds in a Ridge Regression-like penalty and better accounts for cases with high correlated features. It's different for each problem. This means the model fit by lasso regression will produce smaller test errors than the model fit by least squares regression. Let’s first understand the cost function Cost function is the amount of damage you […] Now, linearity is not a standalone property. Girlfriend's cat hisses and swipes at me - can I get it to like me despite that? Ridge regression is an extension of linear regression where the loss function is modified to minimize the complexity of the model. while learning predictive modeling. Ridge Regression is a technique used when the data suffers from multicollinearity ( independent variables are highly correlated). Lasso Regression vs. Ridge Regression. You also need to make sure that the number of features is less than the number of observations before using Ridge Regression because it does not drop features and in that case may lead to bad predictions. Ridge regression adds just enough bias to our estimates through lambda to make these estimates closer to the actual population value. Just as ridge regression can be interpreted as linear regression for which | {
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actual population value. Just as ridge regression can be interpreted as linear regression for which the coefficients have been assigned normal prior distributions, lasso can be interpreted as linear regression for which the coefficients have Laplace prior distributions. Code for this example can be found here. In general, the method provides improved efficiency in parameter estimation problems in … This is added to least square term in order to shrink the parameter to have a very low variance. Linear Regression is so vanilla it hurts. Use MathJax to format equations. Linear Regression establishes a relationship between dependent variable (Y) and one or more independent variables (X) using a best fit straight line (also known as regression line… The Ridge Regression method was one of the most popular methods before the LASSO method came about. Loss function = OLS + alpha * summation (squared coefficient values) It performs better in cases where there may be high multi-colinearity, or high correlation between certain features. The Ridge Regression method was one of the most popular methods before the LASSO method came about. Ridge regression is an extension for linear regression. The Ridge Regression also aims to lower the sizes of the coefficients to avoid over-fitting, but it does not drop any of the coefficients to zero. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. How to gzip 100 GB files faster with high compression. Linear Regression vs. rev 2020.12.10.38158, The best answers are voted up and rise to the top, Data Science Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The canonical example when explaining gradient descent is linear regression. Asking for help, | {
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us. The canonical example when explaining gradient descent is linear regression. Asking for help, clarification, or responding to other answers. •The assumptions of this regression is same as least squared regression except normality is not to be assumed In short, Linear Regression is a model with high variance. The loss function is not really linear in any of its terms, right? In this way, it is also a form of filtering your features and you end up with a model that is simpler and more interpretable. It only takes a minute to sign up. Ridge regression is a regularization technique, which is used to reduce the complexity of the model. The λ parameter is a scalar that should be learned as well, using a method called cross validation that will be discussed in another post. It brings us the power to use the raw data as a tool and perform predictive and prescriptive data… The LASSO method aims to produce a model that has high accuracy and only uses a subset of the original features. Ridge regression Contrast to principal component regression Let contain the 1st k principal components. ”, you agree to our estimates through lambda to make these estimates closer to the potentially high of... Enough bias to the linear relationship among dependent and independent variable whereas it okey. Small \alpha the ridge regression is a model with high variance two or both.... The word the '' in sentences above set of points produce smaller test errors than observations! ( d ) are trivial we discussed l regularization technique, which it is if! Reduces variance in Exchange for bias seen above, if the penalty lambda on the ridge regression vs linear regression, the. Methods before the LASSO method came about that adds a regularization penalty to the potentially high of... Company for its weights despite that that penalty term model that has accuracy! Speakers notice when non-native speakers skip the word the '' in sentences more, see our tips on great... To our estimates through lambda | {
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skip the word the '' in sentences more, see our tips on great... To our estimates through lambda to make these estimates closer to the regression parameters using the values that RSS... Implement ridge regression method without regularization ( penalty on weights ) in Exchange for bias sure if the loss is... The potentially high number of features regression gives an estimate which minimises the sum the. Rss reader model with high variance values of another layer with QGIS expressions a! Than OLS solution, through a better compromise between bias and low variance regression that adds regularization... Requires an initial estimate that is equivalent to the variance norm is called ridge regression is used to overcome in... For salted caramel with matcha really linear in any of its terms, right of... Least squares, and we run into the problems we discussed finance and investing.Linear regression is a technique used the. Let ’ s first understand what exactly ridge regularization:: in regression., linear regression work 2020 Stack Exchange model and use a final model to make predictions for new.... Help, clarification, or responding to other answers this RSS feed, and... Same after reading your answer lambda on the contrary, in the book-editing process can you a. Constraint to the linear regression is a model that has high accuracy and only a! Offers ridge regression is to fit a line to a set of points regression tends to least... They both have cases where there may be high multi-colinearity, or responding to answers... For high school students SGD to solve a problem with a line a non differentiable function. Make these estimates closer to the square of the squares of the squares of the magnitude of original. Overfit the data suffers from multicollinearity ( independent variables are highly correlated ) when non-native speakers skip word. A regularized linear regression is modified to minimize the complexity of the coefficients using the values that minimize.... By a | {
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In loss calculation ( linear regression cost function becomes similar ridge regression vs linear regression the least... First, as others have pointed out, ridge regression is a little different that one could use two! ( also known as Tikhonov regularization ) is a better predictor than least squares fitting procedure the. Prediction error can occur due to variance regression solves the multicollinearity problem through shrinkage parameter λ ( lambda.... Them is whether the model does not overfit the data suffers from multicollinearity ( independent variables highly! A classic a l regularization technique widely used in finance and investing.Linear regression is the classic and simplest! Teenage girls who were interviewed annually for 5 years beginning in 1979 a little different is a common statistical used. May be high multi-colinearity, or high correlation between certain features ridge regression vs linear regression called! Lambda ) λ is high then we will get high bias and low variance a term! Collinear data ( collinearity refers to the linear regression loss function we saw the for! This URL into your RSS reader pointed out, ridge regression b ) and ( )... Data are from the National Longitudinal Study of Youth ( NLSY ) serve a NEMA 10-30 socket dryer... It is one of the features are highly correlated ) order to shrink parameter... Loss function during training estimate that is not a classifier penalty can be adjusted to implement ridge regression different! Multicollinearity in count data analysis, such as negative binomial regression known modeling technique example when gradient... Biased and second is due to any one of the LS coefficients for... Specialized to analyze multiple regression data which is equal to the square of the features to determine where loss! Have cases where they perform better improves the efficiency, but the model is penalized its! L2 regularization or ridge regression is an extension of linear regression, cost! Statistical method used | {
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or ridge regression is an extension of linear regression, cost! Statistical method used in Statistics and Machine Learning here, we saw the equation for linear )... Used … ridge regression is the classic and the simplest linear method of performing regression tasks rotational energy! Square error classic a l regularization technique, which it is here, we saw the equation for regression! Were a large variety of models that learn which features best contribute to the actual population value for ridge is. Is no reason the loss function learned about making linear regression and the simplest linear method of regression. Or personal experience the ridge regression contrast to principal component regression let contain the 1st k principal.... The word the '' ridge regression vs linear regression sentences relationship among dependent and independent variable whereas it is here, the... Magnitude of the coefficients below a fixed value people pick while Learning predictive modeling short, regression... And nature of the coefficients and it helps to reduce the complexity of the LS.... Power to use the raw data as a tool and perform predictive and data…! Very small \alpha the ridge regression: in ridge regression contrast to principal component regression let contain 1st! Regression model where the loss function data set has 1151 teenage girls who interviewed. Data set has 1151 teenage girls who were interviewed annually for 5 years beginning in.... Judge Dredd story involving use of a device that stops time for theft, such negative! Shrink the parameter to have a very low variance a non differentiable loss function can be correlated each. Gzip 100 GB files faster with high compression problems we discussed LASSO are two important regression models and there a. Variables are highly correlated ) regression then fits: the least squares estimate gives: this gives:.. For high school students regularization, if λ is high then we will get high and! A problem with a line to a set of | {
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students regularization, if λ is high then we will get high and! A problem with a line to a set of points with a non differentiable loss function training... About making linear regression using L1 norm is called ridge regression and regression! Of service, privacy policy and cookie policy tricky ( look at a subset of the LS coefficients files with... Before the LASSO and ridge regression model where the multicollinearity is occurring there another vector-based proof for high school?... Among dependent and independent variable can be correlated with each other to other answers regression LASSO., such as negative binomial regression L2 penalty term a tendency to move quickly past vanilla search... Into the problems we discussed have both translational and rotational kinetic energy better in cases they. Regression vs least squares estimate gives: this equation also has an error term is.. Of features any one of these two or both components interpretable due to the.. Looking at a subset of the coefficient that ridge regression, we saw the equation is the linear.... The L2 term is equal to the accuracy of the coefficients the square of original... ( look at a Kronecker product ) fit by LASSO regression will smaller. For logistic regression contributing an answer to data Science Stack Exchange needs to be standardized basic comparison of most... Reduces the standard errors first understand what exactly ridge regularization: its weights we will get high bias low. Use a final model to make these estimates closer to the accuracy of the features to determine where the function... Going to give a basic comparison of the features to determine where the loss function is the regression... The linear least squares estimate gives: this gives: I.e these or! Is linear or non-linear with respect to something treble keys should I have for accordion vs squares! | {
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# Is the set $\left\{x \in Q\colon x^2 \le 2\right\}$ open or closed?
Is the following set open or closed? I am almost certain it is open as the limits are not included in the rational set.
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\}$$
What I really don’t understand is the proper closure of the set. I think it would be:
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\} \cup \left\{\pm\sqrt{2}\right\}$$
But then again, the following seems reasonable (although not an “efficient” closure):
$$\left\{x \in \mathbb{R}\colon x^2 \le 2\right\} = \left[-\sqrt{2},\sqrt{2}\right]$$
• A set if open as a subset of a set with a topology. Which one is the super set with topology that is supposed to contains those sets? $R$? $Q$? – OR. Jun 18 '17 at 2:31
• Take the first set. As a subset of $Q$, considering $Q$ with the topology induced by $R$, we get that the set is open, since $A:=\{x\in Q:\ x^2\leq2\}=Q\cap(-\sqrt{2},\sqrt{2})$. The right-hand side is an open set of $Q$ by definition. On the other hand, as a subset of $R$ the set $A$ is neither open nor closed. It is not closed, as you were saying for having limit points outside itself. It it not open because no neighborhood of $0$ is completely contained in $A$. – OR. Jun 18 '17 at 2:41
• "I am almost certain it is open as the limits are not included in the rational set." That would only mean it is not closed. It wouldn't mean it is open. It could very well be neither. – fleablood Jun 18 '17 at 21:03
• "Is the following set open or closed?" Why do you assume it has to be one or the other? Could it be both? Could it be neither? – fleablood Jun 18 '17 at 21:04
Let $A$ be your set of rationals. You ask if $A$ is open or closed.
Open or closed within what space? The reals $\mathbb{R}$ or the rationals $\mathbb{Q}$?
Notice that $$A = [-r,r] \cap \mathbb{Q} = (-r,r) \cap \mathbb{Q}$$ where r = sqrt 2. Thus within $\mathbb{Q}$, $A$ is clopen.
However, within $\mathbb{R}$ it is neither. | {
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The $\mathbb{R}$-closure of $A$ is $[-r,r]$. The $\mathbb{R}$-interior of $A$ is empty.
It depends on what space your topology is.
If your space is $\mathbb R$ it is neither open nor closed.
If your space $\mathbb Q$ it is both open and closed.
This set is simply $[-\sqrt{2}, \sqrt{2}]\cap \mathbb Q$.
The limit points are all the points between $-\sqrt{2}$ and $\sqrt{2}$. In the space $\mathbb Q$ that is indeed the set. So in the space $\mathbb Q$ this set is closed.
But in the space $\mathbb R$ this includes all the irrational points as well. The rational limit points are included but the irrational limit points are not. So in the space $\mathbb R$ this set is not closed.
In the space $\mathbb Q$ every point is an interior point. Around every point you can find a small enough neighborhood that is entirely inside the set. So every point is an interior point. So the set is open in $\mathbb Q$.
But in the space $\mathbb R$ every neighborhood no matter how small will include irrational points that are not part of the set. So no point of the set is actually an interior point. So the set is not open in $\mathbb R$.
===
In $\mathbb Q$ the set is closed. So the closure is the set itself.
In $\mathbb R$ the set is not closed. To get the proper closure you must include all the limt points. $\pm \sqrt{2}$ are limit points, yes, but so is every point between $-\sqrt{2}$ and $\sqrt{2}$. A limit point isn't just "the points on the edge". There are the points so that every neighborhood no matter how small will intersect the set. These are not only the points "on the edge" but the points "inside" as well.
So the closure will include all the points between $-\sqrt{2}$ and $\sqrt{2}$ and so the closure in $\mathbb Q$ is the set $[-\sqrt{2}, \sqrt{-1}]$.
===
Note:
Open $\ne$ not closed. Sets can be both and sets can be neither.
limit points $\ne$ boundary points. boundary points $\subset$ limit points. | {
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limit points $\ne$ boundary points. boundary points $\subset$ limit points.
It is open in $\Bbb Q$ as you correctly observe. It is closed in $\Bbb Q$ because it is the inverse image of a closed interval under the (continuous) squaring function. | {
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# How can I describe a linear equation that becomes sinusoidal after a certain point?
For example, take $$y = x$$, but at $$x \geq 5$$, $$y$$ now equals $$\sin x + 5$$. So for all $$x$$ values $$\lt 5$$, $$y$$ has linear relationship, but at $$\geq 5$$, $$y$$ now has sinusoidal relationship. Is there a concise way to express this, perhaps a single equation?
A function like this is said to be a piecewise function. The typical way to notate them is to enumerate the subdomains and the functions defined on such:
$$f(x) = \left\{\begin{array}{ll} x, & \text{if }x<5\\ x+\sin x, & \text{if }x\geq 5\\\end{array}\right.$$
Since any real $$x$$ satisfies one of these two conditions, we could also write this as
$$f(x) = \left\{\begin{array}{ll} x+\sin x, & \text{if }x\geq 5\\ x, & \text{else }\end{array}\right.$$
Another, less common notation is that of the Iverson bracket. The Iverson bracket $$[P]$$ for a conditional $$P$$ is defined as $$[P]=\left\{\begin{array}{ll} 1, & \text{if }P\text{ is true}\\ 0, & \text{else }\end{array}\right.$$
This allows us to express piecewise functions with one-line notation, e.g., $$f(x)=x [x<5]+(x\sin x)[x\geq 5].$$ Since $$x\leq 5$$ is the negation of $$x<5$$, we may eliminate the first bracket as $$[x<5]=1-[x\geq 5]$$ and therefore also have \begin{align} f(x)&=x(1-[x\geq 5])+(x+\sin x)[x\geq 5]\\ &=x+(\sin x)[x\geq 5]. \end{align} This captures the intuition that, whether or not $$x\geq 5$$, we start "at minimum" with $$f(x)=x$$; if we $$x\geq 5$$, we add on $$\sin x$$ as well. Of course, we could also have done this with our original notation:
\begin{align} f(x) &= x+ \left\{\begin{array}{ll} \sin x, & \text{if }x\geq 5\\ 0, & \text{else }\end{array}\right.\\ &= x+ (\sin x)\left\{\begin{array}{ll} 1, & \text{if }x\geq 5\\ 0, & \text{else }\end{array}\right. \end{align}
So these are ultimately just variations in notation, all of which present the same concept. | {
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So these are ultimately just variations in notation, all of which present the same concept.
This is called a PIECEWISE function. It is a function that behave differently depending on which parts of the domain you are on. It is basically made out of PIECES of other functions, and they are tied together by taking sections of them and combining them into one function. For example, $$|x|$$ is a piecewise function because it is equal to $$x$$ for $$x \ge 0$$, and it is equal to $$-x$$ for $$x \le 0$$. A piecewise function might look like this: $$f(n) = \begin{cases} n/2, & \text{if n is even} \\ 3n+1, & \text{if n is odd} \end{cases}$$
This question makes me think of a related but slightly different kind of function. A function which gradually/slowly changes behavior, instead of having this piecewise phenomenon.
The piecewise and discontinuous transition would be denoted as $$h(x) = \begin{cases} f(x)\quad\text{if}\quad x \leq x_0\\ g(x)\quad\text{if}\quad x > x_0\\ \end{cases}$$ A gradual shift from $$f$$ to $$g$$ can be written as $$(1-t)\cdot f + t\cdot g$$ with $$t\in[0,1]$$. Therefore a function that gradually changes from $$f$$ after $$x_0$$ and becomes $$g$$ at $$x_1 > x_0$$ would be written $$h_{\mathrm{grad}}(x) = \begin{cases} f(x)\quad\text{if}\quad x \leq x_0\\ \\ \left(1 - \dfrac{x-x_0}{x_1-x_0}\right)\cdot f(x) + \left(\dfrac{x-x_0}{x_1-x_0}\right)\cdot g(x) \quad\text{if}\quad x_0 < x \leq x_1\\ \\ g(x)\quad\text{if}\quad x > x_1\\ \end{cases}$$ As an exemple, let us take $$f(x)=x$$ and $$g(x) = \sin(x) + 5$$ on the domain $$[-5, 10]$$.
The piecewise transition is pictured below:
The continuous/gradual transition is in green below: | {
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# Compute the Number of Combinations
A function to calculate the number of combinations for creating subsequences of k elements out of a sequence with n elements.
from mlxtend.math import num_combinations
## Overview
Combinations are selections of items from a collection regardless of the order in which they appear (in contrast to permutations). For example, let's consider a combination of 3 elements (k=3) from a collection of 5 elements (n=5):
• collection: {1, 2, 3, 4, 5}
• combination 1a: {1, 3, 5}
• combination 1b: {1, 5, 3}
• combination 1c: {3, 5, 1}
• ...
• combination 2: {1, 3, 4}
In the example above the combinations 1a, 1b, and 1c, are the "same combination" and counted as "1 possible way to combine items 1, 3, and 5" -- in combinations, the order does not matter.
The number of ways to combine elements (without replacement) from a collection with size n into subsets of size k is computed via the binomial coefficient ("n choose k"):
To compute the number of combinations with replacement, the following, alternative equation is used ("n multichoose k"):
## Example 1 - Compute the number of combinations
from mlxtend.math import num_combinations
c = num_combinations(n=20, k=8, with_replacement=False)
print('Number of ways to combine 20 elements'
' into 8 subelements: %d' % c)
Number of ways to combine 20 elements into 8 subelements: 125970
from mlxtend.math import num_combinations
c = num_combinations(n=20, k=8, with_replacement=True)
print('Number of ways to combine 20 elements'
' into 8 subelements (with replacement): %d' % c)
Number of ways to combine 20 elements into 8 subelements (with replacement): 2220075
## Example 2 - A progress tracking use-case
It is often quite useful to track the progress of a computational expensive tasks to estimate its runtime. Here, the num_combination function can be used to compute the maximum number of loops of a combinations iterable from itertools: | {
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import itertools
import sys
import time
from mlxtend.math import num_combinations
items = {1, 2, 3, 4, 5, 6, 7, 8}
max_iter = num_combinations(n=len(items), k=3,
with_replacement=False)
for idx, i in enumerate(itertools.combinations(items, r=3)):
# do some computation with itemset i
time.sleep(0.1)
sys.stdout.write('\rProgress: %d/%d' % (idx + 1, max_iter))
sys.stdout.flush()
Progress: 56/56
## API
num_combinations(n, k, with_replacement=False)
Function to calculate the number of possible combinations.
Parameters
• n : int
Total number of items.
• k : int
Number of elements of the target itemset.
• with_replacement : bool (default: False)
Allows repeated elements if True.
Returns
• comb : int
Number of possible combinations.
Examples
For usage examples, please see http://rasbt.github.io/mlxtend/user_guide/math/num_combinations/ | {
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# Number of regions formed by $n$ points in general position
Given $$n$$ points in $$\mathbb{R}^d$$ in general position, where $$n\geq d+1$$. For every $$d$$ points, form the hyperplane defined by these $$d$$ points. These hyperplanes cut $$\mathbb{R}^d$$ into several regions. My questions are:
(1) is there a formula in terms of $$d$$ and $$n$$ that describes the number of regions?
(2) the same question for the number of bounded regions?
I tried many key words on google but found nothing helpful. Any reference or ideas will be appreciated. Thanks
• General position as defined in en.wikipedia.org/wiki/General_position Feb 19, 2020 at 4:27
• Can you maybe elaborate? Both for the general position arrangement where all 4 points lie on the boundary of their convex hull, as well as for the arrangement where one point lies in the interior, I count 6 bounded regions. Feb 19, 2020 at 7:29
• The maximum number of cells in an arrangement of $k$ hyperplanes in dimension $d$ is $O(k^d)$. Your $k$ is $\binom{n}{d}$. Feb 19, 2020 at 11:56
• @JosephO'Rourke Thanks for the info. Can you direct me to a reference where I can find the results? Feb 19, 2020 at 17:59
• @GHfromMO I think you are right, the number of bounded regions may vary from case to case. I am wondering whether the total number of regions would also vary or just be the same? Feb 19, 2020 at 18:01
This is more of a long comment than an answer. It should be possible to compute the number of regions and number of bounded regions using Whitney's theorem for the characteristic polynomial $$\chi(t)$$ (Theorem 2.4 of these notes), and Zaslavsky's theorem that the number of regions is $$(-1)^d \chi(-1)$$, and the number of bounded regions is (in this situation) $$(-1)^d\chi(1)$$ (Theorem 2.5 of the previous link). We need more than the usual definition of "general position." We want the position to be generic enough for the argument below (generalized to $$d$$ dimensions) to hold. | {
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Here is the computation for $$d=2$$. First, the empty intersection (the ambient space $$\mathbb{R}^2$$) contributes $$t^2$$ to $$\chi(t)$$. The $${n\choose 2}$$ lines will contribute $$-{n\choose 2}t$$. Now we must consider all subsets of the lines that intersect in a point $$p$$. Let $$p$$ be one of the original $$n$$ points. Then $${n-1\choose 2}$$ pairs of lines intersect in $$p$$, $${n-1\choose 3}$$ triple of lines intersect in $$p$$, etc., giving a contribution to $$\chi(t)$$ of $${n-1\choose 2}-{n-1\choose 3}+{n-1\choose 4} -\cdots = n-2.$$ We have to multiply this by $$n$$ since there are $$n$$ choices for $$p$$. There are now $$3{n\choose 4}$$ choices of two lines that don't intersect in one of the original $$n$$ points, but they still intersect by genericity. Thus we get an additional contribution of $$3{n\choose 4}$$. It follows that $$\chi(t) = t^2-{n\choose 2}t+n(n-2)+3{n\choose 4}.$$ The number of regions is $$\chi(-1) = \frac 18(n-1)(n^3-5n^2+18n-8).$$ The number of bounded regions is $$\chi(1) = \frac 18(n-1)(n-2)(n^2-3n+4).$$ Can someone extend this argument to $$d$$ dimensions?
Addendum. I worked out $$d=3$$. Here are the details. Let $$X$$ be an $$n$$-element "generic" subset of $$\mathbb{R}^3$$. Thus $$X$$ determines a set $$\mathcal{A}$$ of $${n\choose 3}$$ hyperplanes. We need to find all subsets of $$\mathcal{A}$$ with nonempty intersection. A $$j$$-element subset that intersects in an $$e$$-dimensional affine space contributes $$(-1)^jt^e$$ to the characteristic polynomial $$\chi(t)$$.
Case 1: $$e=3$$. We take the intersection over the empty set to get $$\mathbb{R}^3$$. This gives a term $$t^3$$.
Case 2: $$e=2$$. Each hyperplane contributes $$-t^2$$, giving a term $$-{n\choose 3}t^2$$.
Case 3: $$e=1$$. (a) Any two hyperplanes intersect in a line (by genericity), giving $${{n\choose 3}\choose 2}t$$. | {
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(b) Any $$j\geq 3$$ hyperplanes containing the same two points $$p,q\in X$$ meet in a line. There are $${n\choose 2}$$ choices for $$p,q$$ and $${n-2\choose j}$$ for the remaining element of $$X$$ in the hyperplanes. Thus we get a contribution $${n\choose 2}\sum_{j\geq 3}(-1)^j {n-2\choose j}t = {n\choose 2}\left[-1+(n-2)-{n-2\choose 3}\right]t.$$
Case 4: $$e=0$$. (a) Any three hyperplanes intersect at a point, except when all three contain the same two points $$p,q\in X$$. There are $${{n\choose 3}\choose 3}$$ ways to choose three hyperplanes, and $${n\choose 2}{n-2\choose 3}$$ ways to choose them so that they intersect in two points of $$X$$. Hence we get a contribution $$-\left[ {{n\choose 3}\choose 3}-{n\choose 2}{n-2\choose 3} \right]$$ to the constant term of $$\chi(t)$$ (the minus sign because the number of hyperplanes is odd).
(b) Any $$j\geq 4$$ hyperplanes meeting at a point $$p\in X$$. We can choose $$p$$ in $$n$$ ways. We then must choose $$j$$ two-element subsets of $$X-p$$ whose intersection is empty. There are $${{n-1\choose 2}\choose j}$$ ways to choose $$j$$ two-element subsets of $$X-p$$. If their intersection is nonempty, then they have a common element $$q$$ which can be chosen in $$n-1$$ ways, and then we can choose the remaining elements in $${n-2\choose j}$$ ways. This gives the contribution $$n\sum_{j\geq 4}(-1)^j\left[ {{n-1\choose 2}\choose j}- (n-1){n-2\choose j}\right]$$ $$\ = n\left[ -1+{n-1\choose 2}-{{n-1\choose 2}\choose 2} + {{n-1\choose 2}\choose 3}-(n-1)\left(-1+(n-2) -{n-2\choose 2}+{n-2\choose 3}\right)\right].$$ | {
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(c) Any $$j\geq 3$$ hyperplanes meeting at $$p,q\in X$$, together with one additional hyperplane not containing $$p$$ or $$q$$. There are $${n\choose 2}$$ choices for $$p,q$$ and $${n-2\choose j}$$ ways to choose $$j$$ hyperplanes containing $$p,q$$. There are then $${n-2\choose 3}$$ ways to choose the additional hyperplane not containing $$p$$ or $$q$$. Thus we get the contribution $$-\left[ \sum_{j\geq 3}(-1)^j{n-2\choose j}\right] {n-2\choose 3}$$ $$-{n-2\choose 3}{n\choose 2}\left[-1+(n-2)-{n-2\choose 2} \right].$$
Putting all this together gives the characteristic polynomial $$t^3-{n\choose 3}t^2+ \frac{1}{72}n(n-1)(n-3)(n^3-2n^2-16n+68)t$$ $$-\frac{1}{1296}n(n-2)(n-3)(n^6-4n^5-74n^4+698n^3-2129n^2 +2276n-120).$$ The number of regions is $$\frac{1}{1296}(n-2)(n^8-7n^7-62n^6+938n^5-4295n^4+8429n^3 -4932n^2-2016n-648).$$ The number of bounded regions is $$\frac{1}{1296}(n-1)(n-2)(n-3)(n^6-3n^5-77n^4+603n^3 -1508n^2+1056n+216).$$ Conceivably there could be an error in the computation, but I checked it for $$n=4,5$$ by a brute force computation.
This method should extend to any $$d$$, but the computation will be more complicated, and I am too lazy to work out the details. | {
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• A note on the stronger condition required, here is an example of what can go wrong: Consider the simple case $6$ points in the plane, no $3$ on a line. There is a special degenerate case where we can partition the $6$ points into three pairs $(A, A')$ $(B, B')$ and $(C,C')$ where the lines $AA'$, $BB'$, and $CC'$ all intersect at a point. This will have one less region than the general case.
– Nate
Feb 19, 2020 at 20:54
• Theorem 2.2 of Ardila-Billey arxiv.org/abs/math/0605598 gives a combinatorial description of the relevant matroid (described dually, as the intersections of generic hyperplanes rather than the spans of generic points). I don't see an easy way to extract the characteristic polynomial from it, but maybe you do. Feb 20, 2020 at 4:32
• I replicated your result for $d=2$ based on my answer below. I proved there that $x$ generic lines produce $2x$ unbounded regions and $(x-2)(x-1)/2$ bounded ones. Then replaced $x$ with $n(n-1)/2$. Last, at each of the $n$ points the bounded regions that would exist in a generic $(n-1)$-lines case are collapsed to a point. So the total number of regions collapsed to a point is $n(n-3)(n-2)/2$. This leaves $(x-2)(x-1)/2$-$n(n-3)(n-2)/2$ bounded regions, which matches your calculation. I think $d=3$ could be tacked in a similar way, but much more messy. Feb 29, 2020 at 19:48
Perhaps it is worth quoting this theorem, even though it does not distinguish bounded from unbounded cells, and is phrased in terms of the number of hyperplanes rather than the number of points determining the hyperplanes.
Theorem. Let $$H$$ be a set of $$n$$ hyperplanes in $$\mathbb{R}^d$$. The maximum number of $$k$$-dimensional cells in the arrangement $${\cal A}(H)$$ formed by $$H$$, for $$0 \le k \le d$$, is $$\sum_{i=0}^k \binom{d-i}{k-i} \binom{n}{d-i} \;.$$ The maximum is attained exactly when $${\cal A}(H)$$ is simple. | {
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An arrangement is simple if every $$d$$ hyperplanes meet in a point, and no $$d+1$$ hyperplanes have a point in common. In the OP's situation, the arrangement is non-simple. For example, in $$d=2$$, $$4$$ points determine $$6$$ lines, but each point has $$3$$ lines through it.
For $$d=2$$ and $$k=2$$, the above equation reduces to the familiar expression $$\binom{n}{2} + \binom{n}{1} + \binom{n}{0} = \frac{1}{2}( n^2 + n + 2) \;.$$
Handbook of Discrete and Computational Geometry, 3rd ed. Chapman and Hall/CRC, 2017. CRC link. Chapter 28, Thm.28.1.1, p.724.
This non-answer completes Joseph O'Rourke's nice non-answer, for the case of $$n$$ hyperplanes in $$\mathbb{R}^d$$ in general position. But it also suggests that the OP situation may also well have unique answers.
Define:
$$U_{d,n}=$$ number of unbounded regions cut by $$n$$ hyperplanes in $$\mathbb{R}^d$$
$$B_{d,n}=$$ number of bounded regions
$$T_{d,n}=$$ total number of regions $$=U_{d,n}+B_{d,n}$$
$$S_{d,n}=$$ number of regions cut on the sphere $$S^d$$ by $$n$$ great $$S^{d-1}$$-circles
Then $$U_{d,n}, B_{d,n}$$, $$T_{d,n}$$ and $$S_{d,n}$$ are unique with these formulas:
$$U_{d,0}=1$$, $$\quad B_{d,0}=0$$, $$\quad T_{d,0}=1$$, $$\quad S_{d,0}=1$$
$$U_{1,n}=2$$, $$\quad B_{1,n}=n-1$$, $$\quad T_{1,n}=n+1$$, $$\quad S_{1,n}=2n$$
and for $$n>0$$
1. $$U_{d+1,n}=S_{d,n}$$
2. $$S_{d,n}=U_{d,n}+2B_{d,n}$$
3. $$T_{d,n+1}=T_{d,n}+\sum_{i=0}^{i=d-1}{n\choose i}$$
Proof.
1. In $$\mathbb{R}^{d+1}$$ take a huge and growing $$S^d$$ sphere, so that all the bounded regions zoom down to a point at the center of the sphere, the hyperplanes become great circles on the sphere and the unbounded regions corresponds to regions cut by the circles on the sphere. Therefore if the numbers are unique (as will be proved at the end) 1. follows. | {
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2. Centrally project $$\mathbb{R}^d$$ onto a half $$S^d$$ (tangent to it). Complete the semisphere to a sphere by central symmetry. Then the hyperplanes become great circles, the $$B_{d,n}$$ bounded regions in $$\mathbb{R}^d$$ become $$2B_{d,n}$$ regions in $$S^d$$ and the $$U_{d,n}$$ unbounded ones become $$U_{d,n}$$ regions stretching across the suture line (equator) of the sphere. Again by unicity 2. follows.
3. Start with $$\mathbb{R}^d$$ and $$n$$ hyperplanes in generic position inside it. Now add a new hyperplane in generic position the following way: first chose a point inside one region: no matter how that point is eventually stretched to a hyperplane, to it will split the region in two, for a gain of 1, or $$n \choose 0$$. Now stretch that point to a line: since it is a generic line it will meet each of the $$n$$ hyperplanes once and at each meeting the line will cross into one one new region and split it - with a gain of $$n \choose 1$$ new regions. Next stretch the line to a generic 2-plane, which will meet once each of the $$(d-2)$$-dimensional intersections of two hyperplanes; at each meeting the growing plane will arrive from having already crossed 3 of the 4 regions, to cross into the fourth and cut it; this a gain of another $$n \choose 2$$ regions. In general as a generic $$m-1$$-plane grows to a generic $$m$$-plane it will meet all the $$n \choose m$$ $$(n-m)$$-dimensional intersections of $$m$$ hyperplanes, and each time it will go from cutting $$2^m-1$$ regions before crossing the intersection to cutting all $$2^m$$ after crossing, for a total gain of $$n \choose m$$ regions. This continues up to $$m=d-1$$, proving 3.
Proof of Unicity.
By induction:
$$U_{d,0}$$, $$\quad B_{d,0}$$, $$\quad T_{d,0}$$, $$\quad S_{d,0}$$ are unique;
$$T_{d,n}$$ unique $$\implies$$ $$T_{d,n+1}$$ unique (by the proof of 3.); | {
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$$T_{d,n}$$ unique $$\implies$$ $$T_{d,n+1}$$ unique (by the proof of 3.);
$$U_{d,n}$$ and $$B_{d,n}$$ unique $$\implies$$ $$S_{d,n}$$ unique (by the proof of 2. and the fact that the construction can be reversed in a non-unique way to show that $$S_{d,n}=U_{d,n}+2B_{d,n}$$ for some values of $$U_{d,n}$$ and $$B_{d,n}$$);
$$S_{d,n}$$ unique $$\implies$$ $$U_{d+1,n}$$ unique (by the proof of 1.);
$$U_{d+1,n}$$ and $$T_{d+1,n}$$ unique $$\implies$$ $$B_{d+1,n}$$ unique (as $$B=T-U$$).
• I have a feeling that jiggling the $n \choose d$ hyperplanes of the OP into generic positions while carefully keeping track of the corresponding increase of regions (both bounded and unbounded) one may be able to reach an answer to the OP's question too. For a start this should be relatively easy for $d=2$ (but I haven't done it). Feb 23, 2020 at 15:32 | {
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# Proving the identity $\sum\limits_{k=1}^n {k^3} = {\Large(}\sum\limits_{k=1}^n k{\Large)}^2$ without induction
I recently proved that $$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$ Using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.
Thanks!
-
Look at this takayaiwamoto.com/Sums_and_Series/sumcube_1.html – pritam Jun 27 '12 at 9:29
See math.stackexchange.com/questions/120674 for remarks about proofs "not using induction". – sdcvvc Jun 27 '12 at 10:13
I merged the three existing posts which covered exactly this question, as each post had different interesting answers which should not be lost. I also deleted redundant comments, and comments about closing posts as duplicates. This fourth question is not considered a duplicate. – Eric Naslund Jul 2 '12 at 11:30
Since this question is asked frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does use induction. – Eric Naslund Aug 30 '12 at 0:23
show 1 more comment
Stare at the following image, taken from this MO answer, long enough:
-
+1 I'm in love with that one – leonbloy Sep 2 '11 at 23:23
The fact that there are $k$ blocks (or $\frac{1}{2}+k{-}1+\frac{1}{2}$ blocks) of $k\times k$ size is based on the fact that $\sum\limits_{j=1}^{k-1}=k(k{-}1)/2$. That is, $(k{-}1)/2$ blocks on top $(k{-}1)/2$ on the left and $1$ block at the corner (totaling to $k$). Perhaps I am being picky or slow, but I don't see this as obvious from the image. Beyond that, it is a nice proof-without-words. – robjohn Sep 3 '11 at 4:13
I have put the details of the proof corresponding to this image in this answer since the comment area was too small. – Wok Sep 3 '11 at 7:23
I don't know if this is intuitive, but it is graphic. | {
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I don't know if this is intuitive, but it is graphic.
On the outer edge of each $(k{+}1){\times}k$ block there are $k$ pairs of products each of which total to $k^2$. Thus, the outer edge sums to $k^3$, and the sum of the whole array is therefore $\sum\limits_{k=1}^n k^3$.
The array is the matrix product $$\left[\begin{array}{r}0\\1\\2\\\vdots\\n\end{array}\right]\bullet\left[\begin{array}{rrrrr}1&2&3&\cdots&n\end{array}\right]$$ Therefore, the sum of the elements of the array is $\sum\limits_{k=0}^nk\;\sum\limits_{k=1}^nk=\left(\sum\limits_{k=1}^nk\right)^2$.
Therefore, $\sum\limits_{k=1}^n k^3=\left(\sum\limits_{k=1}^nk\right)^2$
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If we forget the first line of the matrix (which is zero and which is only used to make pairs with the diagonal coefficients), then I like the fact we can put this answer in parallel with the coloured rectangles above and below, and we get another partition of each colored area (each coefficient of the matrix gives a rectangle of a certain area), which explains why each L-shaped area is $k^3$. – Wok Sep 4 '11 at 7:39
However, each L-shape coefficient has the same factor $k$, which means it proves "each L-shaped area is $k^3$" by the same proof that $2 \times \sum_{j=0}^k j = (0+k) + (1+k-1) + ... + (k-1+1) + (k+0) = (k+1) \times k$, which makes it really close to the coloured rectangles above and below. – Wok Sep 4 '11 at 7:49
I hope you don't mind if I use both ideas in another answer. – Wok Sep 4 '11 at 8:13
I don't mind. I simply find it less aesthetic to need to use $\sum\limits_{j=1}^k\;j=k(k+1)/2$ or that $(k(k+1)/2)^2-(k(k-1)/2)^2=k^3$ in an intuitive proof. – robjohn Sep 4 '11 at 19:31
Can you get the intuition explanation from the following two pictures?[EDIT: the following is essentially the same as Mariano's answer. He didn't mentioned the first picture though.]
The images are from Brian R Sears.
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nice, but already posted (linked) by Mariano – leonbloy Sep 2 '11 at 23:24 | {
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nice, but already posted (linked) by Mariano – leonbloy Sep 2 '11 at 23:24
There's this nice picture from the Wikipedia entry on the squared triangular number:
The left side shows that $1 + 2 + 3$ forms a triangle and so that squaring it produces a larger triangle made up of $1+2+3$ copies of the original triangle. The right side has $1(1^2) + 2(2^2) + 3(3^2) = 1^3 + 2^3 + 3^3$. The coloring shows why the two sides are equal.
There are several other references for combinatorial proofs and geometric arguments on the Wikipedia page.
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Each colored area is $k^3$ as a difference of two areas: $S_k^2 - S_{k-1}^2$.
The detailed proof which comes with the drawing is the following.
For any positive integer $k$, we define: $$S_i = \sum_{j=1}^{i} j$$
We first notice: $$S_i^2 = S_i^2 - S_0^2= \sum_{k=1}^{i} \left(S_k^2 - S_{k-1}^2\right)$$
The expected result finally comes from: $$S_k^2 - S_{k-1}^2 = k \left(k+2 S_{k-1}\right) = k\left(k+k\left(k-1\right)\right)=k^3$$
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So essentially, you are using the fact that $$\left(\sum_{j=1}^k\;j\right)^2-\left(\sum_{j=1}^{k-1}\;j\right)^2=k^3$$ to justify the diagram which is supposed to prove that fact intuitively. – robjohn Sep 4 '11 at 1:09
As you mentioned in another answer, this is where the diagram is the least intuitive. However, if you cannot picture $k^3$ on a 2D-plane, then you need another representation as a difference of areas. – Wok Sep 4 '11 at 6:50
As soon as you know $S_k = \sum_{j=1}^k j = \frac{k(k+1)}{2}$, I find it more intuitive to figure out each colored area is $k^3$ on this diagram than to figure it out by counting squares plus two rectangles when $k$ is even. – Wok Sep 4 '11 at 7:06
The formula is due to Nicomachus of Gerasa. There is a nice discussion of ways to prove it at this n-category cafe post, including a bijective proof and some visual / "geometric" proofs.
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Here's another version of this "proof without words". This is the case $n=4$. | {
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Here's another version of this "proof without words". This is the case $n=4$.
There are 1 $1 \times 1$, 2 $2 \times 2$, 3 $3 \times 3$, ... squares, for a total area of $1^3 + 2^3 + \ldots + n^3$. For even $k$, two of the $k \times k$ squares overlap in a $k/2 \times k/2$ square, but this just balances out a $k/2 \times k/2$ square that is left out, so the total is the area of a square of side $1 + 2 + \ldots + n$.
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I found this but sadly do not understand it, can someone explain the method?
Edit:
This has the same proof but much easier to understand. It is also found in the link by QY
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Excellent picture!! (second one) – Eric Naslund Jan 22 '11 at 20:14
Several visual proofs of this indentity are collected in the book Roger B. Nelsen: Proofs without Words starting from p.84.
Although several of these proofs can still be considered inductive, I thought it might be interesting to mention them.
Original sources are given on p. 147:
• 84 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 199. jstor
• 85 Mathematics Magazine, vol. 50, no. 2 (March 1977), p. 74. jstor
• 86 Mathematics Magazine, vol. 58, no. 1 (Jan. 1985), p. 11. jstor
• 87 Mathematics Magazine, vol. 62, no. 4 (Oct. 1989), p. 259. jstor
• 87 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 200. jstor
• 88 Mathematics Magazine, vol. 63, no. 3 (June 1990), p. 178. jstor
• 89 Mathematics Magazine, vol. 62, no. 5 (Dec. 1989), p. 323. jstor
• 90 Mathematics Magazine, vol. 65, no. 3 (June 1992), p. 185. jstor
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Great collections. Thanks – B. S. Jun 27 '12 at 11:11
I believe this illustration is due to Anders Kaseorg:
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We know that $$A=\Sigma_{1}^{n}k^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2$$ and $$B=\Sigma_{1}^{n}k=\frac{1}{2}n^2+\frac{1}{2}n$$ $A-B^2=0$. :) | {
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If you are presenting this proof,write it nicely. $A=\frac{n^2(n+1)^2}{4}$ and $B=\frac{n(n+1)}{2}$,obviously $A=B^2$ – Aang Jun 27 '12 at 10:40
@avatar: Dear Avatar; I will. Your proof is great and it looks analytically. Thanks. :-) – B. S. Jun 27 '12 at 10:48
sorry, if you felt that i was being rude. – Aang Jun 27 '12 at 10:53
@avatar: Your proof lights mine. No Problem at all. WELCOME Avatar. :) :) – B. S. Jun 27 '12 at 10:57
The sum of a degree $n$ polynomial $f(n)$ will be a degree $n+1$ polynomial $S(n)$ for $n \geq 0$ and both polynomials can be extended (maintaining the relation $S(n)-S(n-1) = f(n)$) to negative $n$. To verify that the formula for $\Sigma k^3$ is correct one need only test it for any 5 distinct values of $n$, but the structure of the answer can be predicted algebraically using the continuation to negative $n$.
If $S(n) = (1^3 + 2^3 + \dots n^3)$ is the polynomial that satisfies $S(n)-S(n-1) = n^3$ and $S(1)=1$, then one can calculate from that equation that $S(0)=S(-1)=0$ and $S(-n-1)=S(n)$ for all negative $n$, so that $S$ is symmetric around $-1/2$. The vanishing at 0 and -1 implies that $S(t)$ is divisible as a polynomial by $t(t+1)$. The symmetry implies that $S(t)$ is a function (necessarily a polynomial) of $t(t+1)$.
$S(t)$ being of degree 4, this means $S(n) = a (n)(n+1) + b((n^2 +n)^2$ for constants $a$ and $b$. Summation being analogous to integration (and equal to it in a suitable limit), they have to agree on highest degree terms. Here it forces $b$ to be $1/4$ to match $\int x^3 = x^4/4$. Computing the sum at a single point such as $n=1$ determines $a$, which is zero.
Similar reasoning shows that $S_k(n)$ is divisible as a polynomial by $n(n+1)$ for all $k$. For odd $k$, $S_k(n)$ is a polynomial in $n(n+1)$.
- | {
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You know, $\sum_0^n x^k=\frac{1-x^{n+1}}{1-x}$. Differentiate both sides once, $\sum_1^n kx^{k-1}=\frac{x^n(nx-n-1)+1}{x^2-2x+1}$. Now taking $\lim_{x\to1}$ both sides and then squaring the result will give you the expression on the RHS. You can further differentiate $\sum_0^n x^k=\frac{1-x^{n+1}}{1-x}$ until you get $k^3$ inside the expression, take limit again you will get the same result as of $(\lim_{x\to1}\frac{x^n(nx-n-1)+1}{x^2-2x+1})^2$. You can also prove it using telescopic series.
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why does the assumption hold? this is usually proved using induction... – akkkk Jun 27 '12 at 9:47
what assumption?? – Aang Jun 27 '12 at 9:48
I guess Auke means $\sum_{0}^{n} x^k=\frac{1-x^{n+1}}{1-x}$. – sdcvvc Jun 27 '12 at 10:12
LHS is a geometric series. en.wikipedia.org/wiki/Geometric_progression – Aang Jun 27 '12 at 10:14
@Auke: one can also prove it like this - let $f(x) = \sum\limits_{k=0}^nx^k$, then $f(x) - xf(x)$ is: \begin{align} 1+&x+x^2+\dots+x^n \\ &- \\ &x+x^2+\dots+x^n+x^{n+1} \end{align} which is $1-x^{n+1}$. Hence $$(1-x)f(x) = 1-x^{n+1}$$ as needed. – Ilya Jun 27 '12 at 12:40
http://en.wikipedia.org/wiki/Faulhaber%27s_formula#Faulhaber_polynomials
If $p$ is odd, then $1^p+2^p+3^p+\cdots+n^p$ is a polynomial function of $a=1+2+3+\cdots+n$. If $p=3$, then then the sum is $a^2$; if $p=5$ then it's $(4a^3-a^2)/3$, and so on.
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And $a$ is always a factor of $p$. – lhf Sep 3 '11 at 13:26
http://blogs.mathworks.com/loren/2010/03/04/nichomachuss-theorem/
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$f(n)=1^3+2^3+3^3+......+n^3$
$f(n-1)=1^3+2^3+3^3+......+(n-1)^3$
$f(n)-f(n-1)=n^3$
if $f(n)= (1+2+3+4+....+n)^2$ then
$$f(n)-f(n-1)=(1+2+3+4+....+n)^2-(1+2+3+4+....+(n-1))^2$$
using $a^2-b^2=(a+b)(a-b)$
$f(n)-f(n-1)=$
$=[1+1+2+2+3+3+4+4+....+(n-1)+(n-1)+n][1-1+2-2+3-3+4-4+....+(n-1)-(n-1)+n]=$
$=[2(1+2+3+4+....+(n-1))+n]n=(2\frac{n(n-1)}{2}+n)n=(n(n-1)+n)n=n^3$ | {
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$=[2(1+2+3+4+....+(n-1))+n]n=(2\frac{n(n-1)}{2}+n)n=(n(n-1)+n)n=n^3$
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This is about the same proof as here, the presentation is a bit different though. This is another way to make $k^3$ appear than what was shown here, here and here. | {
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### Author Topic: Q6 TUT 0202 (Read 2820 times)
#### Victor Ivrii
• Elder Member
• Posts: 2599
• Karma: 0
##### Q6 TUT 0202
« on: November 17, 2018, 04:09:19 PM »
Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$f(z)=\frac{1}{e^z-1};\qquad z_0=0\quad \text{(four terms of the Laurent series)} .$$
#### Meng Wu | {
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• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: Q6 TUT 0202
« Reply #1 on: November 17, 2018, 04:09:29 PM »
$$\because e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots$$
\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
$$\\$$
$$\\$$
The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
$$\\$$
$$\\$$ | {
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$$\\$$
$$\\$$
If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
$\\$
Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.
« Last Edit: November 17, 2018, 11:00:48 PM by Meng Wu » | {
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#### Heng Kan
• Full Member
• Posts: 15
• Karma: 18
##### Re: Q6 TUT 0202
« Reply #2 on: November 17, 2018, 05:30:38 PM »
I think you got the coefficients correctly but the Laurent series is wrong. Here is my answer. See the attatched scanned picture.
#### Chunjing Zhang
• Newbie
• Posts: 1
• Karma: 0
##### Re: Q6 TUT 0202
« Reply #3 on: November 17, 2018, 06:00:55 PM »
I think maybe it is needed to first calculate the singularity of the original function, here pole of order 1, and then set the expansion start at power of -1.
#### Victor Ivrii
We have two solutions based on two different ideas: undetermined coefficients and a clever substitution of the power expansion into power expansion. Why clever? Because it chips out $1$ from $(e^z-1)/z$ rather than from $e^z-1$ (which would be an error) | {
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# Proving recursive formula via induction leads to extra term?
I have been asked the following question, and despite spending the last 30 minutes on it, have not come up with a good result:
Define $$f(1) = 2$$, and $$f(n) = f(n-1) + 2n$$ for all $$n \geq 2$$. Find a non-recursive formula for $$n$$, and prove by induction this formula works over all natural numbers.
So, this didn't sound too hard. I found the non-recursive formula to be $$f(n) = n^2 + n$$.
Base case of induction with $$n = 2$$:
$$f(2) = 2^2 + 2 = 6$$
Assumption step:
$$f(k) = k^2 + k$$
Extension step:
\begin{aligned} f(k+1) & = (k+1)^2 + (k+1) \\ & = (k^2 + 2k + 1) + (k+1) \\ & = f(k) + 2k + 2 \ (?) && (\text{Substitute} \ f(k)) \end{aligned}
Notice it all falls apart here? That extra "$$2$$" terms means any experimental result I plug in yields a value $$2$$ higher than it should.
What have I done wrong?
• You're almost done. By the recursive formula, what does $f(k+1)$ equal? Edit: It doesn't fall apart at all. Aug 10 '15 at 10:52
• The recursive formula would be f(k+1) = f(k) + 2(k+1)... ohh I see. I'm an idiot. Why did I miss that? Aug 10 '15 at 10:56
• @EchoLogic, happens to all of us sometimes. Aug 13 '15 at 18:30
• Please add your own answer to this question, completing the details of the above comments, and later accept it to close this question. Sep 3 '15 at 17:01
We have, the recursion formula, \begin{aligned} f(1) = 2, f(n) = f(n-1) + 2n && (1)\end{aligned}
and, the candidate, \begin{aligned}P(n): f(n) = n^2 + n && (2)\end{aligned}
### Step-1: (Base Case)
By $$(1)$$, we have:
$$f(1) = 2$$, and $$f(2) = f(1) + 2 \cdot 2 = 6$$
These are satisfied by $$(2)$$ also, via direct calculation. Hence, the base case is established.
### Step-2: (Induction Hypothesis/ Extension step)
Assume that the statement holds for $$n=k$$,
i.e. $$P(k): f(k) = k^2 + k$$, holds.
### Step-3: (Inductive step)
We now, prove $$P(k+1)$$, which is the subject of the question,
By $$(1)$$, | {
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We now, prove $$P(k+1)$$, which is the subject of the question,
By $$(1)$$,
\begin{aligned} f(k+1) & = f(k) + 2 \cot (k+1) && (\text{Taking,} \ \ n= (k+1)) \\ & = (k^2 + k) + (2k + 2) && (\text{By the Induction Hypothesis}) \\ & = (k^2 + 2k + 1) + (k+1) \\ \implies f(k+1) & = (k+1)^2 + (k+1) \end{aligned}
Thus, $$P(k+1)$$ is established, contingent upon $$P(k)$$ holding.
By the Principle of Mathematical Induction, the formula $$(2)$$ holds.
## Notes:
A. The working presented in the OP has two flaws:
1. The failure to recognize that the expression in the last step has the formula was the desired result, minus taking a common factor. This had been pointed out by @Git Gud and recognized by the OP, in the comments.
2. The more important issue is that the presented working confounds the statement that is given to be true, eq. $$(1)$$, and the one that is to be proved, eq. $$(2)$$. Specifically, the induction hypothesis assumes $$P(k)$$, so that the next step should be to establish $$P(k+1)$$. However, the OP assumes $$P(k+1)$$ in accordance with $$(2)$$, and tries to show how this implies $$(1)$$. Now, this works for the present case as all the operations involved are bidirectional $$(\iff)$$. Thus,we do indeed show that $$(1)$$ implies $$(2)$$, establishing $$P(k+1)$$. However, this is poor form that disrupts the flow of the logic and can lead to trouble if one is dealing with problems regarding say, inequalities. Although working backwards is a legitimate strategy that can be very useful, it requires care to ensure that the reversed steps are valid and useful, and is, at any rate not required here. | {
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B. The notation $$P(k)$$, indicates the statement, $$P(n)$$, that is, ($$f(n) =n^2+n$$), for the case $$n=k$$. $$P(n)$$ is the statement that in the general case that there is an arbitrary $$n \in \mathbb{N}$$, we have $$f(n) = n^2 + n$$. This is a standard notation (albeit not one mentioned in OP's working in the question) in problems where mathematical induction is used, among others.
C. While mathematical induction can be useful for proving true statements over $$\mathbb{N}$$ (and many other cases, including well-founded sets in general), it is not nearly as useful in detecting the invalidity of false statements, and does not the address the important problem of generating plausible statements to check. Indeed, generating the formula is the first part of the problem the OP set out to solve. In the case, the formula was (almost certainly by design), trivial to guess. However, the generation of such statements. and particularly finding explicit formulae for recurrence relations; can be a very challenging problem to solve, and offers an interesting counterpoint to the problem of solving differential equations over $$\mathbb{R}$$ or $$\mathbb{C}$$. While a full discussion would be wildly out of scope here, it is interesting to note that here: $$\frac{f(n)-f(n-1)}{n-(n-1)} = 2n$$ which is reminiscent of the derivative over $$\mathbb{R}$$ and suggesting the appropriate solution should be a quadratic polynomial, as is indeed the case. It is left to the reader to verify that modifying the initial condition leads to solutions of the form $$n^2 + n + f(1) - 2$$. | {
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# A nice way to factorize FRACTIONS!
I recently learned that ...
$\frac{1}{n(n+d)}$ = $\frac{1}{d}(\frac{1}{n}-\frac{1}{n+d})$
For example,
$\frac{1}{3*5}=\frac{1}{3(3+2)}=\frac{1}{2}(\frac{1}{3}-\frac{1}{5})$
Which is useful because we can use this equation in situation like this...
Find the value of $\frac{1}{3*5}+\frac{1}{5*7}+\frac{1}{7*9}+\frac{1}{9*11}$
$=\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{9})+\frac{1}{2}(\frac{1}{9}-\frac{1}{11})$
Take the common factor out
$\frac{1}{2}[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{11})]$
The terms cancels each other and we are left with the first and last terms.
$\frac{1}{2}[\frac{1}{3}-\frac{1}{11}]$
which is $\frac{4}{33}$ This way is MUCH faster than the traditional way
Anyone can prove why the equation i learned is true?
Note by Peter Bishop
5 years, 3 months ago
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$\dfrac{1}{n(n+d)}$
$= \dfrac{1}{d} \times \dfrac{d}{n(n+d)}$
$= \dfrac{1}{d} \times \dfrac{(n+d) - (n)}{n(n+d)}$ | {
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$= \dfrac{1}{d} \times \dfrac{d}{n(n+d)}$
$= \dfrac{1}{d} \times \dfrac{(n+d) - (n)}{n(n+d)}$
$= \dfrac{1}{d} \times \left(\dfrac{n+d}{n(n+d)} - \dfrac{n}{n(n+d)} \right)$
$= \dfrac{1}{d} \times \left( \dfrac{1}{n} - \dfrac{1}{n+d} \right)$
- 5 years, 3 months ago
By using Partial Fractions,let - $\frac{1}{n(n+d)}$ = $\frac{A}{n} + \frac{B}{n+d}$
Thus, we need to find the coefficients A and B. Therefore,by taking LCM,we get -
$\frac{1}{n(n+d)}$ = $\frac{A(n+d) + B \cdot n}{n(n+d)}$. Or, $1= n(A + B) + A \cdot d$
Now, comparing the terms for different exponents of 'n' on both sides, we get-
{ This can be explained by example given below -
If ax + b = 3x +2 , then , a = 3 and b = 2 }
(Comparing the terms which contains $n^0$on both sides),
$n^0 : A \cdot d =1, or , A = \frac{1}{d}$
(Comparing the terms which contains $n^1$ on both sides),
$n^1 : A + B =0 , or, B = -A ,$
Or, $B = \frac{-1}{d}$
Therefore, $\frac{1}{n(n+d)}$ = $\frac{\frac{1}{d}}{n} + \frac{\frac{-1}{d}}{n+d}$
$\frac{1}{n(n+d)}$ = $\frac{1}{d} ( \frac{1}{n} - \frac{1}{n+d})$
- 5 years, 3 months ago
Awesome!!!!!!!!!!!!!!! Thank you very much.
- 5 years, 3 months ago
But i still don't understand the exponents of n part. Can you please explain futher.
- 5 years, 3 months ago
He is using a property called "comparing coefficients" (or at least that is what I call it)
It states the following: For any two polynomials $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$ and $Q(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots +b_1x+b_0$ such that $P(x)=Q(x)$, then $a_i=b_i$ for all $i=0\to n$.
- 5 years, 3 months ago
It's pretty easy.Multiply the RHS by $n(n+d)$ and you will get $\frac {1}{d}.[(n+d)-n]=1$ And that's it.
- 5 years, 3 months ago
wow...good formula for rememberd
- 5 years, 3 months ago
Woah, cool bro!
- 5 years, 3 months ago
Guys, if you like what you learnt, please reshare it so others can see. Thanks for the comments
- 5 years, 3 months ago
Awesome method! From where did you get this?? | {
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- 5 years, 3 months ago
Awesome method! From where did you get this??
- 5 years, 3 months ago
From a tutor
- 5 years, 3 months ago
when we can write the pattern or series in the form that difference of two numbers and such that it get cancelled it makes the work easier this is called v n method or elemination method
- 5 years, 3 months ago
I just learnt A+B=0 so, A=-B wa0w. R I Newton yet?
- 5 years, 3 months ago
1/n(n+d) =1/d x d/n(n=d) =1/d x (n+d)-(n)/n(n+d) =1/d x {(n+d)/n(n+d-(n)/n(n+d)} =1/d x (1/n-1-n+d) HENCE PROVED
- 5 years, 3 months ago
This method is use to solve many problems and is called telescoping method I think.
- 2 years, 10 months ago | {
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# Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?
I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.
Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?
My proof is kinda simple. Let's work in mod $2$. We'll have:
$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$ but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:
$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory.
Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.
Now is there a combinatorial proof of that fact except what I have in mind?
• A standard problem often classified as combinatorics is that the number of people who shook hands with an odd number of people is even. This one uses much the same idea. – André Nicolas Aug 28 '13 at 22:41
• @AndréNicolas: If you already have the proof in your mind, would you please post a proof of my problem using the handshaking lemma? – user66733 Aug 28 '13 at 22:49
• Frankly, I like your proof better than any of the answers. – asmeurer Aug 29 '13 at 4:48
• Another way of phrasing your own solution: We consider the sum of 100 numbers, 50 odd ones and 50 even ones. That will be an even sum. Negating a number (changing sign from $-$ to $+$ or from $+$ to $-$) does not change its parity (its being odd or even). – Jeppe Stig Nielsen Aug 29 '13 at 9:43
• @AndréNicolas: I haven't chosen the best answer yet because I'm still waiting for your answer based on the handshaking Lemma. If you have such a proof in mind please post it, because I'm really interested in it and it sounds like the most combinatorial solution to me. – user66733 Aug 31 '13 at 2:55 | {
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You can rephrase essentially the same argument in the following terms:
Suppose that there were such a pattern of plus and minus signs. Let $P$ be the set of positive terms, and let $N$ be the set of negative terms together with the number $101$. Then $\sum P-\sum N=0$, so $\sum P=\sum N$, and $\{P,N\}$ is a partition of $\{1,2,\ldots,101\}$ into two sets with equal sum. But $\sum_{k=1}^{101}k=\frac12\cdot101\cdot102=101\cdot51$ is odd, so this is impossible.
Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?
Replacing 100 with $n$ and using Brian M. Scott's solution, we want a partition of $\{1, 2, ..., n+1\}$ into two sets with equal sums.
The sum is $\frac{(n+1)(n+2)}{2}$, and if $n=4k$, this is $(4k+1)(2k+1)$ which is odd and therefore impossible.
If $n = 4k+1$, this is $(2k+1)(4k+3)$ which is also odd, and therefore impossible.
If $n = 4k+2$, this is $(4k+3)(2k+2)$, so it is not ruled out, and each sum must be $(4k+3)(k+1)$.
if $n = 4k+3$, this is $(2k+2)(4k+5)$ which is also not ruled out, and each sum must be $(k+1)(4k+5)$.
Now I'll try to find a solution for the not impossible cases. (I am working these out as I enter them.)
For the $n=4k+2$ case, the sum must be $(4k+3)(k+1) =(4k+4-1)(k+1) =4(k+1)^2-(k+1) =(2k+2)^2-(k+1)$. The square there suggests, to me, the formula for the sum of consecutive odd numbers $1+3+...+(2m-1)=m^2$, so $1+3+...+(4k+3) = (2k+2)^2$. If $k+1$ is odd, remove it from the sum so it is $(2k+2)^2-(k+1)$. If $k+1$ is even, both $1$ and $k$ are odd, so remove them from the sum. In either case, we have the desired partition.
For the $n=4k+3$ case, the sum must be $(4k+5)(k+1) =(4k+4+1)(k+1) =4(k+1)^2+(k+1) =(2k+2)^2+(k+1)$. Again, $1+3+...+(4k+3) = (2k+2)^2$. If $k+1$ is even, add it to the sum so it is $(2k+2)^2+(k+1)$. If $k+1$ is odd, $k+2$ is even, so remove $1$ and add $k+2$ to the sum. In either case, we have the desired partition. | {
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I do not know if these partitions are unique.
If $T_n=n(n+1)/2$ is the $n^{th}$ triangular number, an inductive proof (using $T_n+(n+1)=T_{n+1}$) shows the attainable numbers at step $n$ are $$-T_n,\ -T_n+2,\ \cdots , T_n-2, \ T_n,$$ in particular they all have the same parity as $T_n$. Since $T_{100}=5050$ is even, we see that $101$ cannot be attained in any way by $100$ steps.
addendum: The first triangular number at least $101$ is $T_{14}=105$ (and is odd). This overshoots the goal $101$ by $4$, so if we take the sum $$1+2+3+\cdots+14=105$$ and change the sign on the $2$, we get $101$. Seems this is the only way to get $101$ in 14 steps, and we cannot get it with 13 or fewer since $T_{13}=91$ is the largest with $13$ steps.
Consider both sides modulo $2$. Then the right side is $1$, whereas the left side is $0$ (since it consists of $50$ ones and $50$ zeros). | {
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# Why does probability of 50/50 outcome decrease as we toss a coin more and more?
I have following formula ("n" is number of experiments, "k" is number of successes that we expect, "p" is probability of success, "q" is probability of failure):
I used it to calculate chances of getting strictly 2 heads out of 4 tosses (order doesn't matter), 5 heads of 10 tosses, 10 heads out of 20 tosses.
2 heads out of 4 tosses have around 38% probability.
5 heads out of 10 tosses have around 25% probability.
10 heads out of 20 tosses have around 18% probability.
Looks like trend for me. And this trend is paradoxical, it means that by increasing number of tries we will decrease chance for 50/50 outcome. It seems to be at odds with what I was taught before. I was taught that as we increase number of tosses we will come closer and closer to 50/50 outcome. So if I toss a coin 10 times I can get 1 head and 9 tails, while tossing a coin 1000 times is more likely to give me result much closer to 50/50 outcome. | {
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• What p and what q are you using? – Mariano Suárez-Álvarez Feb 13 '18 at 18:20
• This is because throwing exactly 50% is not likely to happen most of the time, you can sometimes land around the 11/20 mark, or 9/20. These are still very close to 10/20, and on average you are still getting an even distribution. As you increase $n$, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values. – John Doe Feb 13 '18 at 18:21
• Notice that that formula gives you the probability of getting exactly $k$ successes. As $n$ increases, it is harder and harder to have exactly $n/2$ tails. – Mariano Suárez-Álvarez Feb 13 '18 at 18:21
• It's true that the probability of getting exactly 50-50 goes down. But the average deviation of from $.5$ gets smaller and smaller. When $n=10$ it's relatively common to see $.6$ or $.4$ (corresponding to $6/10$ or $4/10$ heads.) For $n=10000000$ it is pretty much impossible for $60\%$ or more of the flips to come up heads. – spaceisdarkgreen Feb 13 '18 at 18:33
• Consider also that for an odd number of coin flips, the probability of exactly 50-50 is always zero – spaceisdarkgreen Feb 13 '18 at 18:36
This is because the more you flip a coin, the less likely it will be that you get heads $exactly$ half of the time. It is true, however, that the more you flip a coin, and then compute
$$\hat{p}=\frac{\text{number of heads}}{\text{number of tosses}}$$
that $\hat{p}$ will $converge$ to $0.5$. Consider flipping a coin $10$ times and obtaining $6$ heads. Then consider flipping a coin $1000$ times and obtaining $526$ heads.
While in the second instance, we are $26$ heads away from a perfect $50/50$, proportionally, we are closer as $0.526$ is closer to $0.5$ than $0.6$. | {
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• Please, add information about spreading out probability to your answer because it's incomplete without it. " As you increase n, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values." – user161005 Feb 13 '18 at 18:41
• @user161005 are you talking about my comment on the question? This is not my answer.. – John Doe Feb 13 '18 at 18:42
• Sorry, confused you with other guy. Silly me. – user161005 Feb 13 '18 at 18:52
As it's been said, the fact that there are more and more possible values is what makes $\hat p=0.5$ less and less probable as exact value ($\hat p =\tfrac{\#heads}{\#toses}$).
Consider, for instance, the probability of $\tfrac 1 2 - \tfrac 1 {10}<\hat p_n<\tfrac 1 2 + \tfrac 1 {10}$, that is $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_n<\tfrac 1 2 + \tfrac 1 {10}\right)=P\left(- \tfrac {2\sqrt n} {10}<\frac{\hat p_n-\tfrac 1 2}{\frac1{2\sqrt n}} < \tfrac {2\sqrt n} {10}\right)\approx 2\cdot\Phi\left(\tfrac {\sqrt n} {5}\right)-1$$ the approximation based on the CLT, valid for $n$ large enough.
Let's supose $n=50$. In that case we have $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_{50}<\tfrac 1 2 + \tfrac 1 {10}\right)\approx 2\cdot \Phi\left(\frac{\sqrt {50}}5\right)-1\approx 0.8427.$$ | {
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Now let's do the same fot $n=100$. We get $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_{100}<\tfrac 1 2 + \tfrac 1 {10}\right)\approx 2\cdot \Phi\left(\frac{\sqrt {100}}5\right)-1\approx 0.9545.$$ So the probability of $\hat p_n$ taking a value at a distance less than $0.1$ from the expected value has increased by around a 13%. But the values of $X=\#heads$ corresponding to that event went from $9$ (that is $21$ to $29$) to $19$ ($41$ to $59$), around the double. And if we kept doubling $n$ (for example), we would approximately double the number of integer values of $X$ that lead to $0.4<\hat p_n<0.6$ each time, but the total probability distributed between those values won't grow that much, fundamentally because it's very close to $1$, which is a bound, no matter $n$.
So, those probabilities which stay almost constant at some point, are distributed among more and more possible values of the measured discrete variable, and so is not surprising that the probability at each individual point decreases. | {
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# Limit of $\frac{1}{x^4}\int_{\sin x}^{x} \arctan t dt$
I am trying to find this limit,
$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$
Using the fundamental theorem of calculus, part 1, $\arctan$ is a continuous function, so $$F(x):=\int_0^x \arctan{t}dt$$ and I can change the limit to $$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$
I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.
I tried using l'Hospital's rule, but the calculation gets tedious.
Can anyone give me hints?
EDIT
I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!
• Numerical evaluation says the limit should be $1/6$. – Simply Beautiful Art Mar 29 '17 at 23:34
• @SimplyBeautifulArt How did you do it? What kind of method did you use for numerical evaluation? – zxcvber Mar 29 '17 at 23:36
• Calculators of course. – Simply Beautiful Art Mar 29 '17 at 23:37
Power series can help, if you know them.
We know that $$\arctan t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+1}}{2n+1},\qquad \lvert t\rvert<1.$$ Therefore an antiderivative for $\arctan t$ is $$F(t):=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+2}}{(2n+1)(2n+2)},\qquad \lvert t\rvert<1.$$ So, as $x\to0$, $$F(x)=\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).$$ Now, recalling that $$\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+O(x^5)=x+O(x^3),$$ we see that as $x\to0$ (and therefore $\sin x\to0$), \begin{align*} F(\sin x)&=\frac{1}{2}\left(x-\frac{x^3}{6}+O(x^5)\right)^2-\frac{1}{12}\left(x+O(x^3)\right)^4+O[(x+O(x^3))^6]\\ &=\frac{1}{2}\left[x^2-\frac{x^4}{3}+O(x^6)\right]-\frac{1}{12}\left[x^4+O(x^6)\right]+O(x^6)\\ &=\frac{x^2}{2}-\frac{x^4}{4}+O(x^6) \end{align*} So, all told, our limit is $$\lim_{x\to0}\frac{F(x)-F(\sin x)}{x^4}=\lim_{x\to0}\frac{\,\frac{x^4}{6}\,}{x^4}=\frac{1}{6}$$ | {
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• I really like the way you used $O(x^n)$ notation here. Thanks! – zxcvber Mar 29 '17 at 23:56
Here's one way to do this. The Taylor expansion for $\arctan(x) = x - x^3/3 + x^5/5 + \cdots$. So then $$\int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12 + x^6/30) - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{12} - \frac{(\sin x)^6}{30}.$$
Using another Taylor expansion, $\sin x = x - x^3/3! + x^5/5! + \cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$), $$(\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$ and $$(\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$ Thus $$\int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$
So multiplying by $x^{-4}$ and taking the limit gives $1/6$.
As an aside, l'Hopital's rule would also work.
Since $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get
$$\lim_{x \to 0}\frac{F(x)-F(\sin x)}{x^4} = \lim_{x \to 0}\frac{\arctan x - \cos x\arctan \sin x}{4x^3}.$$
This last limit can be evaluated using Taylor series:
$$\arctan x = x-\frac{x^3}{3}+O(x^5)$$ and
$$\cos x \arctan \sin x = x-x^3+O(x^5)$$
and the limit you are looking for is equal to
$$\lim_{x \to 0}\frac{x-\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \frac{1}{6}.$$
The ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\sin$, $\cos$ and $\arctan$. Easier done than said.
• Actually, you just take the derivatives directly instead of trying to combine well-known expansions since that's ugly and we only need a few terms. – Simply Beautiful Art Mar 29 '17 at 23:49
• @SimplyBeautifulArt That's true, but I expect the third derivative of $\cos x \arctan \sin x$ to be even uglier.. – Stefano Mar 29 '17 at 23:51
• Well, the key is that its easy to ask your calculator to take the third derivative ;) Combining Taylor expansions? Not so much. – Simply Beautiful Art Mar 29 '17 at 23:53 | {
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Note that by the Mean Value Theorem for integrals we have $$\int_{\sin x}^{x}\arctan t\,dt = (x - \sin x)\arctan c$$ for some $c$ between $x$ and $\sin x$. Then we have $$\lim_{x \to 0}\frac{1}{x^{4}}\int_{\sin x}^{x}\arctan t\,dt = \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{\arctan c}{x}$$ The first factor on the right tends to $1/6$ (via L'Hospital's Rule or Taylor series) and we show that next factor tends to $1$. For this we assume that $x \to 0^{+}$ so that $\sin x < c < x$ and therefore $$\arctan \sin x < \arctan c < \arctan x$$ and dividing by $x$ we get $$\frac{\arctan \sin x}{\sin x}\cdot\frac{\sin x}{x} < \frac{\arctan c}{x} < \frac{\arctan x}{x}$$ By Squeeze theorem we see that $(\arctan c)/x \to 1$ as $x \to 0^{+}$. A similar argument can be given for $x \to 0^{-}$ and we have the desired limit as $1/6$. | {
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Welcome to our community
youth4ever
New member
Hello my Math friends,
I have a complex integration problem:
The integral to calculate is :
Integrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]
or if particularized for a=1, m=1 the integral will be :
Integrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]
I know the answer for both of them:
(Pi/a)*(e^-ma) for the first
and
Pi/e for the second.
How this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.
Thanks.
PS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?
Opalg
MHB Oldtimer
Staff member
Hello my Math friends,
I have a complex integration problem:
The integral to calculate is :
Integrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]
or if particularized for a=1, m=1 the integral will be :
Integrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]
I know the answer for both of them:
(Pi/a)*(e^-ma) for the first
and
Pi/e for the second.
How this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.
Thanks.
Hi youth4ever and welcome to MHB!
Your integral is the real part of $$\displaystyle \int_{-\infty}^\infty \frac{e^{imx}}{x^2+a^2}dx.$$ If you know about contour integration then you should be able to evaluate that integral by using a D-shaped contour.
PS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?
See the LaTeX tips and tutorials forum.
youth4ever
New member
Hi Opalg,
Yes I know that it comes from e^imx/... integral. But I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.
Is this possible ?
Thanks.
ThePerfectHacker | {
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Thanks.
ThePerfectHacker
Well-known member
But I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.
Is this possible ?
I think your question is. Rather than making things "complicated", why not just find anti-derivative of the function inside the integral and get the answer. But the problem with what you want is that a lot of these integration problems that use complex analysis to compute do not have anti-derivatives in terms of known standard functions. So what you want cannot be done.
Random Variable
Well-known member
MHB Math Helper
If your interested in a different approach that would confirm the answer you get using contour integration, consider the Laplace transform of $$\displaystyle f(t) = \int_{0}^{\infty} \frac{\cos (mtx)}{a^{2}+x^{2}} \ dx$$
Then by definition,
$$\mathcal{L}_{t} [f(t)](s) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\cos (mtx)}{a^{2}+x^{2}} \ e^{-st} \ dx \ dt = \int_{0}^{\infty} \frac{1}{a^{2}+x^{2}}\int_{0}^{\infty} \ \cos(mxt) e^{-st} dt \ dx$$
$$= \int_{0}^{\infty} \frac{1}{a^{2}+x^{2}} \frac{s}{s^{2}+(mx)^{2}} \ dx = \frac{s}{s^{2}-a^{2}m^{2}} \int_{0}^{\infty} \left( \frac{1}{a^{2}+x^{2}} - \frac{m^{2}}{s^{2}+m^{2}x^{2}} \right)$$
$$= \frac{s}{s^{2}-a^{2}m^{2}} \left( \frac{\pi}{2a} - \frac{\pi m }{2s} \right) = \frac{\pi}{2a(s+am)}$$
Now undo the transform.
EDIT: What I had previously after this point wasn't quite correct.
$$f(t) = \mathcal{L}^{-1}_{t} \Big[ \mathcal{L}_{t} [f(t)](s) \Big] = \mathcal{L}^{-1}_{t} \Big[ \frac{\pi}{2a(s+am)} \Big]$$
$$= \frac{\pi}{2a} \mathcal{L}^{-1}_{t} \Big[ \frac{1}{s+am} \Big] = \frac{\pi}{2a} e^{-amt}$$
And therefore,
$$\int_{-\infty}^{\infty} \frac{\cos (mx)}{a^{2}+x^{2}} \ dx = 2 f(1) = \frac{\pi}{a} e^{-am}$$
Last edited:
youth4ever
New member
Hello "Random Variable" | {
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