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# Polar moment of inertia

moment

Note: Different disciplines use the term moment of inertia to refer to different moments. In physics, moment of inertia is strictly the second moment of mass with respect to distance from an axis, which characterizes an object's angular acceleration due to an applied torque. In engineering (especially mechanical and civil), moment of inertia commonly refers to the second moment of the area. When reading polar moment of inertia take care to verify that it is referring to "polar second moment of area" and not moment of inertia. Polar second moment of area will have units of length to the fourth power (e.g. ${\displaystyle m^{4}}$ or ${\displaystyle in^{4}}$), while moment of inertia is mass times length squared (e.g. ${\displaystyle kg*m^{2}}$ or ${\displaystyle lb*in^{2}}$).

The polar second moment of area (also referred to as "polar moment of inertia") is a measure of an object's ability to resist torsion as a function of its shape. It is one aspect of the second moment of area linked through the perpendicular axis theorem where the planar second moment of area uses a beam's cross-sectional shape to describe its resistance to deformation (bending) when subjected to a force applied in a plane parallel to its neutral axis, the polar second moment of area uses a beam's cross-sectional shape to describe its resistance to deformation (torsion) when a moment (torque) is applied in a plane perpendicular to the beam's neutral axis. While the planar second moment of area is most often denoted by the letter, ${\displaystyle I}$, the polar second moment of area is most often denoted by either, ${\displaystyle I_{z}}$, or the letter, ${\displaystyle J}$, in engineering textbooks.

The calculated values for the polar second moment of area are most often used describe a solid or hollow cylindrical shaft's resistance to torsion, as in a vehicle's axle or drive shaft. When applied to non-cylindrical beams or shafts, the calculations for the polar second moment of area becomes erroneous due to warping of the shaft/beam. In these instances, a torsional constant should be used, where a correctional constant is added to the value's calculation.

A schematic showing how the polar second moment of area ("Polar Moment of Inertia") is calculated for an arbitrary shape of area, R, about an axis o, where ρ is the radial distance to the element dA.

The polar second moment of area carries the units of length to the fourth power (${\displaystyle L^{4}}$); meters to the fourth power (${\displaystyle m^{4}}$) in the metric unit system, and inches to the fourth power (${\displaystyle in^{4}}$) in the imperial unit system. The mathematical formula for direct calculation is given as a multiple integral over a shape's area, ${\displaystyle R}$, at a distance ${\displaystyle \rho }$ from an arbitrary axis ${\displaystyle O}$.

${\displaystyle J_{O}=\iint \limits _{R}\rho ^{2}dA}$.

In the most simple form, the polar second moment of area is a summation of the two planar second moments of area, ${\displaystyle I_{x}}$ and ${\displaystyle I_{y}}$. Using the Pythagorean theorem, the distance from axis ${\displaystyle O}$, ${\displaystyle \rho }$, can be broken into its ${\displaystyle x}$ and ${\displaystyle y}$ components, and the change in area, ${\displaystyle dA}$, broken into its ${\displaystyle x}$ and ${\displaystyle y}$ components, ${\displaystyle dx}$ and ${\displaystyle dy}$.

Given the two formulas for the planar second moments of area:

${\displaystyle I_{x}=\iint \limits _{R}x^{2}dxdy}$, and ${\displaystyle I_{y}=\iint \limits _{R}y^{2}dxdy}$

The relation to the polar second moment of area can be shown as:

${\displaystyle J_{O}=\iint \limits _{R}\rho ^{2}dA}$

${\displaystyle J_{O}=\iint \limits _{R}(x^{2}+y^{2})dxdy}$

${\displaystyle J_{O}=\iint \limits _{R}x^{2}dxdy+\iint \limits _{R}y^{2}dxdy}$

${\displaystyle \therefore J=I_{x}+I_{y}}$

In essence, as the magnitude of the polar second moment of area increases (i.e. large object cross-sectional shape), more torque will be required to cause a torsional deflection of the object. However, it must be noted that this does not have any bearing on the torsional rigidity provided to an object by its constituent materials; the polar second moment of area is simply rigidity provided to an object by its shape alone. Torsional rigidity provided by material characteristics is known as the shear modulus, ${\displaystyle G}$. Linking these two components of rigidity, one can calculate the angle of twist of a beam, ${\displaystyle \theta }$, using:

${\displaystyle \theta ={\frac {Tl}{JG}}}$

Where ${\displaystyle T}$ is the applied moment (torque) and ${\displaystyle l}$ is the length of the beam. As shown, higher torques and beam lengths lead to higher angular deflections, where higher values for the polar second moment of area, ${\displaystyle J}$, and material shear modulus, ${\displaystyle G}$, reduces the potential for angular deflections.


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• I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
• (+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
• At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
• To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
• PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07

First of all to follows we remember some elementary definitions and results about manifolds.

Definition

A function $$f$$ defined in a subset $$S$$ of $$\Bbb R^k$$ is said of class $$C^r$$ if it can be extended to a function $$\phi$$ (said $$C^r$$-extension) that is of class $$C^r$$ in a open neighborhood of $$S$$.

Lemma

If $$f$$ is a function defined in a subset $$S$$ of $$\Bbb R^n$$ such that for any $$x\in S$$ there exist a function $$f_x$$ defined in a neighborhood of $$x$$ that is of class $$C^r$$ and compatible with $$f$$ on $$U_x\cap S$$ then $$f$$ is of class $$C^r$$.

Lemma

If $$U$$ is an open set of $$H^n_k:=\Bbb R^{n-k}\times[0,+\infty)^k$$ for any $$k\le n$$ then the derivatives of two different extensions $$\phi$$ and $$\varphi$$ of a $$C^r$$-function $$f$$ agree in $$U$$.

Definition

A $$k$$-manifold with boundary/corners in $$\Bbb R^n$$ of calss $$C^r$$ is a subspace $$M$$ of $$\Bbb R^n$$ whose points have a neighborhood $$V$$ in $$M$$ that is the immage of a homeomorphism $$\phi$$ of calss $$C^r$$ defined an open set $$U$$ of $$\Bbb R^k$$ or of $$H^k_1/H^k_m$$ and whose derivative has rank $$k$$.

So now let be $$M$$ a $$(n-1)$$-manifold with boundary of class $$C^r$$ in $$\Bbb R^n$$ and thus let be $$\gamma$$ a injective curve defined in the unit interval $$I:=[0,1]$$ such that $$\gamma(0)=0$$ and such that the unit tangent vector not lies -for each $$t\in I$$- to the tangent space at any point of $$M$$. So we call cylindroid $$C$$ of trajectory $$\gamma$$ and of section $$M$$ the set $$C:=\bigcup_{t\in I}\big(M+\gamma(t)\big)$$ that is obtained moving along $$\gamma$$ the points of $$M$$. Click here to see an example. So if $$\xi\in C$$ then there exist a coordinate patch $$\alpha:U\rightarrow V$$ and $$t\in I$$ such that $$\xi=\alpha(x)+\gamma(t)$$ for any $$x\in U$$ and thus let be $$\phi$$ the function from $$U\times I$$ to $$\Bbb R^n$$ defined through the equation $$\phi(x,t):=\alpha(x)+\gamma(t)$$ for any $$(x,t)\in U\times I$$ and then we prove that any restriction of this function is a coordinate patch about $$\xi$$. So we observe that the immage of $$\phi$$ is open in $$C$$: indeed the immage $$V$$ of $$\alpha$$ is an open set of $$M$$ so that there exist an open set $$W$$ of $$\Bbb R^n$$ whose intersection with $$M$$ is $$V$$ and so remembering that the translation is a homeomorphism we observe that $$\phi[U\times I]=\bigcup_{t\in I}\big(V+\gamma(t)\big)=\bigcup_{t\in I}\Big((W\cap M)+\gamma(t)\Big)=\bigcup_{t\in I}\Big(\big(W+\gamma(t)\big)\cap\big(M+\gamma(t)\big)\Big)=\\ \Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap\Biggl(\bigcup_{t\in I}\big(M\cap\gamma(t)\big)\Biggl)=\Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap C$$ and thus we conclude that the immage of $$\phi$$ is open in $$C$$ because $$W+\gamma(t)$$ is open for each $$t\in I$$. Now if $$\overset{°}M$$ and $$\partial M$$ are the sets of the interior and boundary points of $$M$$ then we observe that $$C$$ is union of the sets $$\bigcup_{t\in\text{int}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in\text{bd}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in I}\big(\partial M+\gamma(t)\big)$$ so that we analyse separately the case where $$\xi$$ is an element of the first set and the case where $$\xi$$ is an element of the second or either of the third: in particular this means to analyse separately the case where the set $$U$$ is open in $$\Bbb R^{n-1}$$ and $$t$$ is an element of $$\text{int} \, I$$ and the case where this is not and so just this is what we will do to follows -clearly this can be done independentely from the definition of the three mentioned sets nevertheless we decided to define them because we thought it makes more clear the following argumentetions, that's all. In particular in the example posted above these sets are respectively the invisible part, the red part, the green and balck parts. So first of all we observe that if $$\xi$$ is an element of the first set then $$\alpha$$ is a coordinate patch of $$M$$ defined in an open set $$U$$ of $$\Bbb R^{n-1}$$ and so the map $$\phi$$ defined above is a diffeomorphism in a neighborhood at any point of $$U\times\text{int}\,I$$ where is contained $$\phi^{-1}(\xi)$$, because if the unit tangent vector of $$\gamma$$ not lies to the tangent space at any point of $$M$$ then the derivative of $$\phi$$ is an isomorphism and so the statement follows directely from the inverse function theorem. So we conclude that the set $$\bigcup_{t\,\in\,\text{int}\,I}\big(\overset{°}M+\gamma(t)\big)$$ is a $$n$$-manifold without boundary. In particular in this way we proved that the invisible part of the linked example is a manifold without boundary. Now if $$\xi$$ is a not an element of the first set then the previous argumentations hold only with some efforts that we show to follow. So the functions $$\alpha$$ and $$\gamma$$ can be extended to two $$C^r$$-functions $$\beta$$ and $$\psi$$ defined in a open neighborhood of $$U$$ and $$I$$ respectively so that the function $$\phi$$ can be extended to a function $$\varphi:=\beta+\psi$$ defined in a open neighborhood of $$U\times I$$ and in particular at any point $$(x,t)$$ of $$U\times I$$ this function has not singular derivative so that by the inverse function theorem there exist a (rectangular) open neighborhood $$W_x\times W_t$$ where $$\varphi$$ is a diffeomorphism. Now the set $$\varphi[W_x\times W_t]$$ is open in $$\Bbb R^n$$ and it is not disjoint from $$C$$ -indeed $$W_x\times W_t$$ is not disjoint from $$U\times I$$ and $$\varphi$$ is compatible with $$\phi$$ in $$U\times I$$ and the immage of $$\phi$$ is contained in $$C$$- so that by the continuity of $$\phi$$ the set $$\phi^{-1}[\varphi[W_x\times W_t]]$$ is (not empty and) open in $$U\times I$$ and contains $$W_x\times W_t\cap U\times I$$ where $$\phi$$ is a diffeomorphism. So if we prove that the immage of $$W_x\times W_t\cap U\times I$$ through $$\phi$$ is open in $$C$$ then we conclude that the restriction of $$\phi$$ to this set is a coordinate patch defined in a open set of $$\Bbb R^{n-1}\times[0,+\infty)$$ or in a open set of $$\Bbb R^{n-2}\times[0,+\infty)^2$$ if $$U$$ is open in $$\Bbb R^{n-1}$$ and $$t$$ is not an element of $$\text{int}\, I$$ or if $$U$$ is not open in $$\Bbb R^{n-1}$$ and $$t$$ is an element of $$\text{int}\, I$$ or either if $$U$$ is not open in $$\Bbb R^{n-1}$$ and $$t$$ is not an element of $$\text{int}\, I$$ respectively. So how do this? Could someone help me, please?

To follows some observations that I tried to use: naturally you are not constrained to read it if you do not desidre.

OBSERVATION

If the function $$\phi$$ was injective in $$U\times I$$ the statement follows immediately because in particular the function $$\varphi$$ would be injective in $$W_x\times W_t\cup U\times I$$ so that $$\phi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t]\cap\varphi[U\times I]=\\ \varphi[W_x\times W_t]\cap\phi[U\times I]$$ having remembered that $$\varphi$$ and $$\phi$$ are compatible in $$U\times I$$. So in particular we observe that the injectivity of $$\phi$$ follows immediately if the set $$M+\gamma(t_0)$$ and $$M+\gamma(t_1)$$ was disjoint for any $$t_0,t_1\in I$$ such that $$t_0\neq t_1$$ and in particular I tried to prove this using the injectivity of $$\gamma$$ that above I did not use. Moreover since $$\alpha$$ is a homeomorphism and since $$W_x$$ is open in $$\Bbb R^{n-1}$$ then $$\alpha[W_x\cap U]$$ is open in $$M$$ and so exist a open set $$W$$ of $$\Bbb R^n$$ whose intersection with $$M$$ is $$\alpha[W_x\cap U]$$ and thus implementing the argumentations used above it follows that $$\phi[W_x\times W_t\cap U\times I]=...=\Biggl[\bigcup_{t\in W_t\cap I}\big(W+\gamma(t)\big)\Biggl]\cap\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl]$$ so that if the set $$\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl]$$ and $$\Biggl[\bigcup_{t\in I\setminus W_t}\big(M+\gamma(t)\big)\Biggl]$$ was disjoint then $$\phi[W_x\times W_t\cap U\times I]$$ was open in $$C$$ and this surely happens if the set $$M+\gamma(t_0)$$ and $$M+\gamma(t_1)$$ was disjoint for any $$t_0,t_1\in I$$ such that $$t_0\neq t_1$$.

Comment My initial thoughts go in the following direction, but by second read I see that this concerns likely only a subset of impossible solutions. But perhaps the ansatz itself helps as initial idea...

Error detected - needs update

(taken from my two earlier comments)
By subtracting $$8$$ on each side one gets $$7⋅(7^X−1)=3⋅(3^Y−1)+5⋅(5^Z−1)$$ and $$(X,Y,Z)=(x−1,y−1,z−1)$$. Then divide all by $$7⋅3⋅5$$.

For instance we get on the lhs $$(7^X−1)/15$$ and on the rhs equivalent terms.

This gives restrictions for the modular classes for the exponents, and possibly one can proceed with further modular considerations (for instance on primefactorizations in lhs and rhs).

For instance the lhs has then always $$2^5⋅5$$ as factor, the terms in the rhs always $$2^4$$ and $$2^3$$ as factor. Since the constant occurence of $$2^5$$ as minimum means a strong selection for the $$5^Z−1$$-term: only $$Z=24j$$ is possible and by this the selected have all the primefactors $$3⋅13⋅31⋅31$$. This can similarly be checked for the other terms and often it is possible to derive from this contradictions.

By the corresponding primefactors of 2 and 13 in the lhs and rhs I find that in the lhs we can only have valid solutions in steps of 12 in the exponent. The primefactor decomposition of the first two exponents give

(7^12k-1)/15:
X,    value,                primefactors
--------------------------------------------------------------
[12, 922752480,            "2^5.3.13 .5.19.43.181 "]
[24, 12772082092037760960, "2^6.3.13 .5.19.43.181  xxx.73.193.409.1201"]

This means, with all following admissible exponents we shall have the set of primefactors of the first row.

Similarly we look at the $$5^z$$ term, in already required $$24$$ steps:

(5^24j-1)/21:
Z    value                primefactors
--------------------------------------------------------------
[24, 2838316417875744,     "2^5.3.13   .31.313.601.390001"]
[48, 16917684184764290..., "2^6.3.13   .31.313.601.390001.17.11489.152587500001"]

Again, in all following admissible exponents we shall have the set of primefactors in the first row. But interesting: $$7^{12k}-1$$ as well as $$5^{24j}-1$$ have the primefactor $$3$$ - but of course this cannot happen in the third term:

(3^24i-1)/35:
Y    value                primefactors
--------------------------------------------------------------
[24, 8069415328,           "2^5  .13      .41.73.6481"]
[48, 227904123076...     , "2^6  .13      .41.73.6481   .17.97.193.577.769"]

So, if I've not made a silly mistake, this is a proof of nonexistence of solutions aside $$(X,Y,Z)=(0,0,0)$$ resp $$(x,y,z)=(1,1,1)$$. This answer space is free to use :)

• @MakotoKato Could you please delete the content of your answer except the remark "... is free to use" - so that other users can use this space without worry? – IV_ Jul 13 at 20:00

$$\frac 1x + \frac 1y + \frac 1z = \frac 56$$

$$\frac 1x + \frac 1y = \frac 56 - \frac 1z$$

$$\frac{x+y}{xy} = \frac {5z-6}{6z}$$

$$\frac{x+y}{xy} = \frac {5z-6}{6z}6 • You should remove "This answer is free for anyone to use" at the beginning. I almost overwrote this. – robjohn Jun 6 at 22:08 I have a simple inversion routine for lower triangular matrices (assumed to be invertible, no errorchecks) in Pari/GP I get, dependent on counting, for multiplications either 1/6*x^3 + 1/2*x^2 + 1/3*x (where x means dimension $$n$$) or 1/6*x^3 - 1/3*x ; see below.  \\ invert a lower triangular matrix; no errorchecks! {triinv(m)=local(tmp,rs=rows(m),cs=cols(m),su_m,su_d); tmp=matrix(rs,cs); su_m=su_d=0; \\ inserted for documenting the multiplications & divisions for(c=1,cs, tmp[c,c]=1/m[c,c]; su_d++; for(r=c+1,rs, tmp[r,c]=-sum(k=c,r-1,m[r,k]*tmp[k,c]) / m[r,r]; su_m+= (r-c); su_d++; ); ); \\ return(m); \\ commented out, we need the counters of mult. & divs! return([cs,su_m,su_d]);}  Document: \\ document the multiplications, divisions in a list for m=1..20 for(m=1,20,print( triinv(PPow(1,m)))); \\ PPow gives a Pascalmatrix of size m x m  Protocol: \\ protocol \\ dim mu di "mu"=multiplications "di"=divisions \\ [1 0 1] \\ [2 1 3] \\ [3 4 6] \\ [4 10 10] \\ [5 20 15] \\ [6 35 21]  Find formula: \\ for multiplications we get the sequence [0,1,4,10,20,...] \\ for divisions [1,3,6,10,...] polinterpolate([1,4,10,20]) \\ check formula for multiplications \\ %7 = 1/6*x^3 + 1/2*x^2 + 1/3*x \\ result by Pari/GP polinterpolate([0,1,4,10,20]) \\ %8 = 1/6*x^3 - 1/6*x \\ result by Pari/GP  This space is now free for anyone to use • @AbhigyanSaha Could you please delete the content of your answer except the remark "This space is now free for anyone to use" - so that other users can use this space without worry? – IV_ Jul 13 at 20:00 For the general case a multistep-transformation, be it a cycle or not, of $$N$$ steps and the sum-of-exponents $$S=A_1+A_2+...+A_N$$ where $$a_{k+1}={ 3a_k+1\over 2^{A_k} }$$ and $$A_k=\nu_2(3a_k+1)$$ might be written as $$a_{N+1}=T(a_1;[A_1,A_2,...,A_N])$$ or - a so to say "canonical form" -: $$a_{N+1}={3^N \over 2^S} a_1 + T(0;[A_1,A_2,...,A_N])$$ The latter form I've once named "canonical form" because here $$T()$$ is free from any variable $$a_1$$ and the constant $$0$$ is inserted for all cases, and the $$T()$$-notation hides your explicite $$\sum()$$-notation of powers of products of $$3$$ and of $$2$$. For example we can take the example $$(N,S)=(3,5)$$ and $$E_N:=[A_1,A_2,A_3]=[1,2,2]$$ then a first solution is via $$T(0;E_N)= {9+3\cdot 2 + 2^3\over 2^5}={23\over 32}$$ $$a_4 = { 27\over 32} a_1 + {23\over 32} \\ \implies \\ a_1= 43 + k\cdot 2 \cdot 2^S \\ a_4= 37 + k\cdot 2 \cdot 3^N$$ Note, that the $$+k \cdot 2 \cdot 2^S$$-term (and following the $$\cdot 3^N$$ term) occur by the modularity condition such that the rhs must be integral. That notation shows modularity to the modulus $$2 \cdot 2^S$$: a second solution $$b_1 \to b_4$$ to such a transformation is in the modulus to $$2 \cdot 2^S$$; such that, for instance, $$b_1 = a_1 + 2 \cdot 2^S$$ and then $$b_{N+1}={3^N \over 2^S} b_1 + T(0;E_N)$$ $$b_{N+1}={3^N \over 2^S} (a_1+2 \cdot 2^S) + T(0;E_N)$$ Then follows $$b_{N+1}=a_{N+1}+ {3^N \over 2^S} (2 \cdot 2^S) \\ b_{N+1}=a_{N+1}+ 2 \cdot 3^N$$ So $$b_1 - a_1 = 2 \cdot 2^S \\ b_{N+1} - a_{N+1} = 2 \cdot 3^N$$ Another view, which is immediately derivable, is that $$2^S \cdot a_{N+1} - 3^N a_1 = 2^S \cdot T(0;E_N) \\ 2^S \cdot b_{N+1} - 3^N b_1 = 2^S \cdot T(0;E_N)$$ Of course we can as well subtract $$2 \cdot 2^S$$ from $$a_1$$ to get $$c_1=a_1-2 \cdot 2^S$$ and then $$c_{N+1}=a_{N+1}- {3^N \over 2^S} (2 \cdot 2^S) = a_{N+1}- 2 \cdot 3^N$$ such that Question title: Can first order logic be extended to include infinite conjunctions? Can first order logic (FOL) be extended in some way where infinite conjunctions are permissible? Specifically, can it be extended and still refute statements in a finite number of steps? I would like this question to be answered using Tarskian semantics, where names refer to objects external to the logic. Self answer: Suppose that an infinite conjunction is legal syntax in a first order theory. In first order logic, we are free to negate any formula we can reason about, so a theory which can reason about infinite disjunctions will necessarily reason about their negation. $$$$\tag{1} \{P_a,P_b,\dots\}$$$$ $$$$\tag{2} \lnot(P_a\lor P_b\lor\dots)$$$$ Consider a first order theory which contains the infinite set of formulae $$(1)$$ and the negation of an infinite disjunction $$(2)$$. Any interpretation which satisfies $$(2)$$ will necessarily make the set of formulae $$(1)$$ unsatisfiable. However, every finite subset of formulae will be satisfiable, which by compactness means that the whole thing is. This is a contradiction, so it must not be the case that an infinite conjunction can be made legal syntax in a first order theory. Question title: What did Hilary Putnam mean by this following quote of his? In Putnam's paper "The logic of quantum mechanics", he states: There is nothing really answering to the Copenhagen idea that two kinds of description (classical and quantum mechanical) must always be used for the description of physical reality (one kind for the ways to be used for the 'observer' and the other for the 'system'), nor to the idea that measurement changes what is measured in a indescribable way (or even brings into existence), nor to the 'quantum potential', 'pilot waves', ect. of the hidden variable theorists. These no more than Reichenbach's 'universal forces'. Precisely what did he mean by this? In particular, what did he mean by the "idea that measurement changes what is measured in a indescribable way (or even brings into existence)"? How does one clearly define the the issue raised in the first point, and has it been resolved today? From this you have bounds for your $$r(a_1)$$-parameter: $$r(a_1) = { E \over N+E } = 1-{ 1 \over E/N+1 } \implies$$ $$E_{min}=1-\{N \gamma_1 \} + N \gamma_1 \\ E_{max}=N$$ $$\displaystyle \lim_{N \to \infty} r(a_1)_{E_{min}} = 1-{ 1 \over (1-\{N \gamma_1 \} + N \gamma_1 )/N+1 } = 1-\frac 1{\gamma_1+1} = 1- \log_3(2) = \log_3(1.5)$$ $$\displaystyle \lim_{N \to \infty} r(a_1)_{E_{max}} = { E_{max} \over N + E_{max} } = { N \over 2N } = \frac 12$$$$

{\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} Let us assume the "Syracuse"-style notation of the Collatz-iteration $$a_{k+1}= {3a_k+1\over 2^{A_1}} \qquad a_k \small \text{ from the odd integers}$$ and for a $$N$$-fold iterated transformation the short, vectorial, notation: $$a_{N+1}=T(a_1;[A_1,A_2,...,A_N])$$ So let $$N$$ denote the (N)umber of steps $$3x+1$$ and $$S$$ denote the (S)um of the exponents $$A_k$$, which is also the number of $$x/2$$-steps. Then, to convert this into the version of $$(3x+1)/2$$ and $$x/2$$ -stepping, we introduce $$E$$ the number of even steps without the $$(3x+1)/2$$ steps, so $$E=S-N$$. With that, I understand your $$r(a_1)$$ as $$r(a_1)=E/(N+E) = (S-N)/S = 1- N/S$$. We can observe, • that the trival cycle $$1 = T(1;[2,2,2,...2])$$ to any length $$N$$, has the values $$S=2N$$ and $$E=N$$ and $$r(a_1)= 1-N/S = 1-N/(2N)=1/2$$ • that the first cycle in the negative numbers $$-1= T(-1;[1,1,1,1,...,1])$$ to any length $$N$$, has $$S=N$$, $$E=0$$, and $$r(a_1)= 0$$ • that the second cycle in the negative numbers $$-5=T(-5;[1,2,1,2,1,2,...,1,2])$$ to any even length $$N=2n$$, has $$S=3n$$, $$E=n$$ and $$r(a_1) = E/(N+E)= n/(3n) = 1/3$$ Now to have a cycle of any length, and other than $$T(a_1;[2,2,2,...,2])$$ we can use the well known multiplication-formula for the $$N$$ members of an expected cycle $$a_k$$ ($$k=1..N$$) $$a_2 \cdot a_3 \cdot ... \cdot a_N \cdot a_1 = \left({3a_1+1\over 2^{A_1}}\right) \left({3a_2+1\over 2^{A_2}}\right) \cdots \left({3a_N+1\over 2^{A_N}}\right)$$ This can be rearranged to $$2^S = 2^{A_1+A_2+...A_N} =\left(3+{1\over a_1}\right) \left(3+{1\over a_2}\right) \cdots \left(3+{1\over a_N}\right)$$ We see, that the rhs must be at least as large as the smallest perfect power of $$2$$ larger than $$3^N$$, but at most as $$4^N = 2^{2N}$$ so we get for the lhs (writing $$\gamma=\log_2(3)$$, and further below $$\gamma_1=\log_2(3)-1$$): $$2^{\lceil N \cdot \gamma \rceil} \le 2^S \le 2^{2N}$$ which in terms of $$S$$ means $$\lceil N \cdot \gamma \rceil \le S \le 2N \qquad \text{where } S \in \mathbb N^+$$ and in terms of $$E$$ instead $$\lceil N \cdot \gamma \rceil -N =\lceil N \cdot \gamma_1 \rceil \le E \le N$$ From this you have bounds for your $$r(a_1)$$-parameter: $$r(a_1) = { E \over N+E } \implies \\ {\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E}$$ Well, this formula, in which $$E$$ must be evaluated after $$N$$ is given and might be checked for calculation of $$r(a_1)$$ between $$\lceil N \gamma_1 \rceil$$ and $$N$$, looks not very nice to me, so I'd reconsider the choice for the ratio-parameter $$r()$$. FREE REAL ESTATE $${}{}{}{}{}$$ $$=\underbrace{4n^2(n-1)(n+1)}_{\equiv ~0~(\text{mod}~~ 48)}+\underbrace{(n-1)n(n+1)(n+2)}_{\equiv ~0~(\text{mod}~ 24)}$$ Feel free to improve my post: (This post is an attempted answer to the question: Has research involving the generalization of the arithmetic mean been done for the following?) # Defining First Measure If $$S\subseteq A$$: • $$\ell$$ is the length of an interval • $$\left(I_{n}\right)_{j=1}^{n}$$ and $$\left(J_{k} \right)_{k=1}^{m}$$, for $$m\in\mathbb{N}$$, are a sequence of open intervals where $$\left(I_1\right)=\left(I_2\right)=...=\left(I_{n}\right)=g\in\mathbb{R}^{+}$$, $$\;\ell(J_1)=...=\ell(J_m)=g\in\mathbb{R}^{+}$$ and the infimum in the equations below are taken over all possible $$I_j$$ and $$J_k$$ \begin{align} \mathcal{M}(g,S)=\begin{cases} g\cdot\inf\left\{m\in\mathbb{N}: {S^{\prime}\text{ is countable}, \; \left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is uncountable}\\ g\cdot\inf\left\{m\in\mathbb{N}: {S\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is countable} \end{cases} \end{align} and $$\mathcal{N}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{M}\left(g,S\right)}{\mathcal{M}\left(g,A\right)}$$ then if \begin{align} \mathcal{O}(g,S)= \small{g\cdot\inf\left\{m\in\mathbb{N}: {S_{j}\subseteq S, \; \mathcal{N}(S_{j},A)=0,\;\bigcup\limits_{j=1}^{\infty}{S}_{j}=S^{\prime},\;\left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\}} \end{align} (I added $$\mathcal{O}(g,S)$$ so when $$A$$ is an intervals of finite lengths, $$\mathcal{O}(g,S)$$ gives the Lebesgue Measure of $$S$$ divided by the Lebesgue Measure of $$A$$.) therefore the outer measure is $$\mathcal{P}^{*}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{O}(g,S)}{\mathcal{O}(g,A)}$$ And the inner measure is $$\mathcal{P}_{*}(S,A)=1-\mathcal{P}^{*}(A\setminus S,A)$$, which means measure $$\mathcal{P}(S,A)$$ exists when $$\mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A)$$. Now note $$S$$ and $$A$$, when $$\mathcal{P}(S,A)$$ is undefined, can also have a measure when $$S$$ and $$A$$ have a algebraic notation written in terms of subsets of $$\mathbb{R}$$. # Defining Algebraic Form For arbitrary set $$A$$, if • $$p_{1,1},...,p_{w,q_{w}}\subseteq\mathbb{R}$$ and $$z_{1,1},...,z_{w,v_w}\subseteq\mathbb{R}$$ (For example the subsets can be $$\mathbb{Z}$$, the Cantor Set, or finite length intervals and:) • $$w=[1,...,l\in\mathbb{N}]$$ • $$T_{1,1},...,T_{l,w}:p_{1,1,1} \times...\times p_{l,w,q_{l,w}}\times z_{1,1},...,z_{w,v_w}\to\mathbb{R}$$ then the algebraic form of $$A$$, or $$\alpha$$ is: \begin{align} & \alpha=\bigcup_{w=1}^{l}\bigcap\limits_{u_{w,1}\in z_{w,1}}...\bigcap\limits_{u_{w,v_{w}}\in z_{w,v_{w}}}\bigcup\limits_{s_{w,1}\in p_{w,1}}...\bigcup\limits_{s_{w,q_{w}}\in p_{w,q_{w}}}\left\{T_{l,w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)\right\} \end{align} To illustrate, when set $$A$$ has an algebraic form $$\alpha_1=\bigcup\limits_{m\in\mathbb{Z}}\bigcup\limits_{n\in\mathbb{Z}}\left\{\frac{m}{8n+4}\right\}$$ where • $$l\in \left(k_1=\left\{1\right\}\right)$$ • $$w\in \left(k_2=\left\{1\right\}\right)$$ • $$s_{1,1,1}=m$$ and $$p_{1,1,1}=\mathbb{Z}$$ • $$s_{1,1,2}=n$$ and $$p_{1,1,2}=\mathbb{Z}$$ other algebraic forms must be equivelant to $$\alpha_{1}$$. For example, another algebraic form of $$A$$ is $$\alpha_2=\bigcup\limits_{m_1\in\mathbb{Z}}\bigcup\limits_{m_2\in\mathbb{Z}}\left\{\frac{2m_1+1}{8m_2+4}\right\}\cup\bigcup\limits_{m_3\in\mathbb{Z}}\bigcup\limits_{m_4\in\mathbb{Z}}\left\{\frac{m_{3}}{12m_4+6}\right\}$$ where • $$w\in k_1=\left\{1\right\}$$ • $$l\in k_2=\left\{1,2\right\}$$ • $$s_{1,1,1}=m_1$$ and $$p_{1,1,1}=\mathbb{Z}$$ • $$s_{1,1,2}=m_2$$ and $$p_{1,1,2}=\mathbb{Z}$$ • $$s_{1,2,1}=m_3$$ and $$p_{1,2,1}=\mathbb{Z}$$ • $$s_{1,2,2}=m_4$$ and $$p_{1,2,2}=\mathbb{Z}$$ Since $$\alpha_1=\alpha_2$$, $$\alpha_2$$ is an algebraic form of $$A$$. Moreover, note we can define the set of all algebraic forms of $$A$$ as $$\mathcal{A}(A)$$ where $$\alpha\in\mathcal{A}(A)$$. For example: $$\mathcal{A}(\mathbb{Q})=\left\{\left\{\frac{m}{n}:m,n\in\mathbb{Z}\right\},\left\{\frac{m}{2n}:m,n\in\mathbb{Z}\right\},...,\left\{\frac{m}{n^2}:m,n\in\mathbb{Z}\right\},...\right\}$$ And an $$\alpha\in\mathcal{A}\left(\mathbb{Q}\right)$$ is $$\alpha=\left\{\frac{m}{n}:m,n\in\mathbb{Z}\right\}$$ Therefore, we can now define a "Folner-like" Sequence inorder to define a measure # Defining New Measure Going back to $$\alpha$$. If • $$\Phi_{1,w}=\left\{1,...,q_{w}\in\mathbb{N}\right\}$$ • $$\Phi_{2,w}=\left\{1,...,v_{w}\in\mathbb{N}\right\}$$ • $$i_{1,w},r_{w}\in\Phi_{{1,w}}$$ • $$i_{2,w}\in\Phi_{2,w}$$ • $$t_{l,w}\in\Phi_{l,w}\setminus\{r_{l,w}\}$$ then $$\alpha$$ becomes \begin{alignat}{2} &\mathcal{F}(w,\alpha,\psi,r_{l,w},\epsilon,\omega)= \\ &\bigcup\limits_{i_{l,w}\in\Phi_{l,w}}\bigcap\limits_{\large{u_{i_{w}}\in z_{i_{w}}}}\bigcup\limits_{\large{s_{w,i_{w}}\in p_{i_{w}}}}&&\left\{T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right):\left|s_{w,t_{w}}\right|\le \omega, |u_{w,i_{w}}|\le\eta\right. \\ & &&\left. \left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon \right\} \end{alignat} Note if all $$p_{w}$$ are subsets of $$\mathbb{Z}$$ and $$\omega\to\infty$$ for every $$\eta$$ when $$\eta\to\infty$$, then each $$s_{r_{w}}$$ would give different Folner Sequences. However, this forces us to choose a particular $$r_{w}$$. Also note we add $$\left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon$$ so when $$A$$ is dense in an interval and $$S\subseteq A$$ is dense almost nowhere, the density of $$S$$ using Folner-esque sequences is always zero. To avoid choosing between different $$r_{w}$$, take the intersection of all of them and take the Folner Sequences of the $$l$$ and $$w$$'s which helps obtain a measure that matches $$\mathcal{P}$$ when $$\mathcal{P}(S,A)$$ is defined. \begin{align} \alpha_{\mathcal{F}}\left(\alpha,\psi,\epsilon,\omega,\eta\right)=\bigcup\limits_{w=1}^{l}\bigcap\limits_{r_{w}\in\Phi_{w}}\mathcal{F}\left(w,\alpha,\psi,r_{w},\epsilon,\omega,\eta\right) \end{align} (Note we take $$k_1$$ less than $$\eta$$ so when $$A=\mathbb{Q}$$ one can derive a unique Folner Sequence which gives a unique measure.) However, there are multiple algebraic forms of $$\alpha$$ in $$A$$ which prevent us from obtaining such a measure. Therefore we add additional criteria: If $$\alpha_1,\alpha_2\in\mathcal{A}(A)$$ we solve $$\beta=\inf\left\{\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)\right\}$$ and take the set of all $$\alpha_1$$ that gives $$\beta$$. $$\mathcal{A}_{\beta}(A)= \left\{\alpha_1:\left(\forall\alpha_{2}\right)\left(\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)=\beta\right)\right\}$$ For example, when $$A=\mathbb{Q}$$, the Folner Sequence $$\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\bigcap_{k\in\mathbb{N},|k|\le\eta}\left(\bigcup\limits_{n\in\mathbb{Z},|n|\le\omega}\bigcup_{m\in\mathbb{Z}}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\cup\bigcup\limits_{n\in\mathbb{Z}}\bigcup_{m\in\mathbb{Z},|m|\le\omega}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\right)$$ Should be in $$\mathcal{A}_{\beta}(A)$$ (We can picture the triple-limit as follows: For every $$\epsilon>0$$ and $$\eta>0$$, take $$\omega$$ as $$\omega\to\infty$$ then for every $$\epsilon>0$$ take $$\eta$$ as $$\eta\to\infty$$. As $$\epsilon$$ is evaluated for smaller values, the limit should converge. Then take the union of all $$\alpha$$ in $$\mathcal{A}_{\beta}(A)$$ to get: $$A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)=\bigcup\limits_{\alpha\in\mathcal{A}_{\beta}(A)}\alpha_{\mathcal{F}}(\alpha,\psi,\epsilon,\omega)$$ (Presumably), we are not only dealing with countable sets. Hence instead of taking the density of $$S\subseteq A$$ using Folner Sequences: \begin{align} \lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\frac{\left|S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|}{\left|A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|} \end{align} We take \begin{align} d(S,\psi)=\lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\mathcal{P}\left(S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right),A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right) \end{align} Now, even if $$d(S,\psi)$$ exists, it only takes the density over $$[\psi-\epsilon,\psi+\epsilon]$$. Therefore, we broaden the density to all $$A$$ by taking the weighted average of $$d(S,\psi)$$ using $$\mathcal{P}$$. In formal terms: Divide $$[0,1]$$ ($$0\le d(S,\psi)\le 1$$) into partitions $$0=c_1\le \cdot\cdot\cdot\le c_i \le \cdot\cdot\cdot\le c_u=1$$ where if $$x_i\in[c_{i-1},c_i]$$ and $$C_i=[c_{i-1},c_i]$$, then our final outer measure is \begin{align} & \mathcal{U}^{*}(S,A)= \lim\limits_{u\to\infty}\sum\limits_{i=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(S,\psi\right)=C_i\right\},A\right) \end{align} And our final inner measure is \begin{align} & \mathcal{U}_{*}(S,A)= 1-\lim\limits_{u\to\infty}\sum\limits_{c=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(A\setminus S,\psi\right)=C_i\right\},A\right) \end{align} Therefore, $$\mathcal{U}(S,A)$$ exists when: $$\mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A)$$ # Generalizing the Measure We can generalize our measure to $$\mu$$: $$$$\mu(S,A)=\begin{cases} \mathcal{P}(S,A) & \mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A) \\ \mathcal{U}(S,A) & \mathcal{P}^{*}(S,A)\neq\mathcal{P}_{*}(S,A), \; \mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A) \end{cases}$$$$ # Defining The Average To define the average, first split $$[a,b]=[\min(A),\max(A)]$$ into sub-intervals using partitions $$x_i$$ $$a= x_0 \le \dots \le x_i \le \dots \le x_r =b$$ Here, if $$1\le i \le r$$, then $$[x_{i-1},x_i]$$ are sub-intervals of $$[a,b]$$. For every $$i$$, choose a $$v_i\in A\cap[x_{i-1}, x_{i}]$$ and define $$A_i=A\cap[x_{i-1},x_i]$$. Next we define set $$P$$, such that $$i \in P\subseteq\left\{1,...,r\in\mathbb{N}\right\}$$ when $$A\cap[x_{i-1},x_i]\neq\emptyset$$. This gives \begin{align} & \lim_{r\to\infty}\sum_{i\in P} f(v_i) \times \mu(A_i,A) \end{align} (This may look tedious but we can use shortcuts to simplify the sum. If I write it out the post will be too long.) \begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y-y)}{x+y} \\ &\,=\lim_{(x,y) \to (0,0)} \dfrac{\,1 - \cos (x+y)\cos y-\sin (x+y)\sin x\,}{x+y} \\ &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y)}{x+y}\\ &= 0.\end{align} 1. To see how any formula was written in any question or answer, including this one, right-click on the expression and choose "Show Math As > TeX Commands". (When you do this, the '' will not display. Make sure you add these. See the next point. Or click the Edit link at the bottom o a post to view the source code. 1. For inline formulas, enclose the formula in .... For displayed formulas, use ...$$. These render differently. For example, type \sum_{i=0}^n i^2  to show $$\sum_{i=0}^n i^2$$ (which is inline mode) or type $$\sum_{i=0}^n i^2
to show $$\sum_{i=0}^n i^2$$ (which is display mode).

2. There are in lower case Greek letters, $$\alpha, \beta, \ldots, \omega$$ and uppercase, $$\Gamma, \Delta, \ldots, \Omega$$. Some Greek letters have variant forms: $$\epsilon$$, $$\varepsilon$$, $$\phi$$, $$\varphi$$ and others.

3. For superscripts and subscripts, use ^ and _. For example $$x_i^2$$, $$\log_2 x$$, $$x_{i,j}$$.

4. Groups. Superscripts, subscripts, and other operations apply only to the next “group”. A “group” is either a single symbol, or any formula surrounded by curly braces {}. If you do 10^10, you will get a surprise: $$10^10$$. But 10^{10} gives what you probably wanted: $$10^{10}$$. Use curly braces to delimit a formula to which a superscript or subscript applies: x^5^6 is an error; {x^y}^z is $${x^y}^z$$, and x^{y^z} is $$x^{y^z}$$. Observe the difference between x_i^2 $$x_i^2$$ and x_{i^2} $$x_{i^2}$$.

5. Parentheses $$(2+3)[4+4]$$,$$\{a,b,c\}$$.

These do not scale with the formula in between, so if you write (\frac{\sqrt x}{y^3}) the parentheses will be too small: $$(\frac{\sqrt x}{y^3})$$. Using \left(\right) will make the sizes adjust automatically to the formula they enclose: \left(\frac{\sqrt x}{y^3}\right) is $$\left(\frac{\sqrt x}{y^3}\right)$$.

\left and\right apply to all the following sorts of parentheses: $$(x)$$, $$[x]$$, $$\{ x \}$$, $$|x|$$, $$\Vert x \Vert$$, $$\langle x \rangle$$, $$\lceil x \rceil$$, and $$\lfloor x \rfloor$$.

6. Sums and integrals the subscript is the lower limit and the superscript is the upper limit, $$\sum_1^n$$. Don't forget {} if the limits are more than a single symbol. For example, \sum_{i=0}^\infty i^2 is $$\sum_{i=0}^\infty i^2$$. Similarly, \prod $$\prod$$, \int $$\int$$, \bigcup $$\bigcup$$, \bigcap $$\bigcap$$, \iint $$\iint$$, \iiint $$\iiint$$, \idotsint $$\idotsint$$.

7. Fractions and binomials $$\frac 17 23 \frac{17}{23}$$,$$\frac ab$$,$$\frac{a+1}{b+1}\binom{n+1}{2k}$$

8. Different Fonts

• $$\mathbb{C}$$, $$\mathbb{R}$$, $$\mathbb{Q}$$, $$\mathbb{Z}$$,
• $$\mathcal{CHNQRZ}$$
• $$\mathscr{CHNQRZ}$$
• $$\mathfrak{CHNQRZ}$$.
1. Radical signs / roots Use sqrt, which adjusts to the size of its argument: \sqrt{x^3} $$\sqrt{x^3}$$; \sqrt[3]{\frac xy} $$\sqrt[3]{\frac xy}$$. For complicated expressions, consider using {...}^{1/2} instead.

2. Some special functions such as $$\sin,\cos, \tan, \cot, \arcsin,\arccos, \arctan$$ $$\sinh,\cosh, \tanh, \coth, \arsinh,\arccos, \sec$$

$$\ln, \log,\lg,\log_2,\log_{16}$$ Use subscripts to attach a notation to \lim: \lim_{x\to 0} $$\lim_{x\to 0}$$ Nonstandard function names can be set with \operatorname{foo}(x) $$\operatorname{foo}(x)$$.

1. There are a very large number of special symbols and notations, too many to list here; see this shorter listing, or this exhaustive listing. Some of the most common include:
• $$\lt$$, $$\gt$$, $$\le$$, $$\leq$$, $$\leqq$$, $$\leqslant$$, $$\ge$$, $$\geq$$, $$\geqq$$, $$\geqslant$$, $$\neq$$. You can use \not to put a slash through almost anything: \not\lt $$\not\lt$$ but it often looks bad.
• $$\times$$, $$\div$$, $$\pm$$, $$\mp$$. \cdot is a centered dot: $$x\cdot y$$
• $$\cup$$, $$\cap$$, $$\setminus$$, $$\subset$$, $$\subseteq$$, $$\subsetneq$$, $$\supset$$, $$\in$$, $$\notin$$, $$\emptyset$$, $$\varnothing$$
• {n+1 \choose 2k} or \binom{n+1}{2k} $${n+1 \choose 2k}$$
• $$\to$$, $$\rightarrow$$, $$\leftarrow$$, $$\Rightarrow$$, $$\Leftarrow$$, $$\mapsto$$
• $$\land$$, $$\lor$$, $$\lnot$$, $$\forall$$, $$\exists$$, $$\top$$, $$\bot$$, $$\vdash$$, $$\vDash$$
• $$\star$$, $$\ast$$, $$\oplus$$, $$\circ$$, $$\bullet$$
• $$\approx$$, $$\sim$$, $$\simeq$$, $$\cong$$, $$\equiv$$, $$\prec$$, $$\lhd$$, $$\therefore$$
• $$\nabla$$, $$\partial$$ \Im \Re $$\Im$$, $$\Re$$
• For modular equivalence, use \pmod like this: a\equiv b\pmod n $$a\equiv b\pmod n$$.
• For the binary mod operator, use \bmod like this: a\bmod 17 $$a\bmod 17$$.
• Avoid using \mod, as it produces extra space: compare the above with a\mod 17 $$a\mod 17$$.
• \ldots is the dots in $$a_1, a_2, \ldots ,a_n$$ \cdots is the dots in $$a_1+a_2+\cdots+a_n$$

Detexify lets you draw a symbol on a web page and then lists the $$\TeX$$ symbols that seem to resemble it. These are not guaranteed to work in MathJax but are a good place to start. To check that a command is supported, note that MathJax.org maintains a list of currently supported $$\LaTeX$$ commands, and one can also check Dr. Carol JVF Burns's page of $$\TeX$$ Commands Available in MathJax.

1. Spaces MathJax usually decides for itself how to space formulas, using a complex set of rules. Putting extra literal spaces into formulas will not change the amount of space MathJax puts in: a␣b and a␣␣␣␣b are both $$a b$$. To add more space, use \, for a thin space $$a\,b$$; \; for a wider space $$a\;b$$. \quad and \qquad are large spaces: $$a\quad b$$, $$a\qquad b$$.

To set plain text, use \text{…}: $$\{x\in s\mid x\text{ is extra large}\}$$. You can nest $…$ inside of \text{…}, for example to access spaces.

1. Accents and diacritical marks Use \hat for a single symbol $$\hat x$$, \widehat for a larger formula $$\widehat{xy}$$. If you make it too wide, it will look silly. Similarly, there are \bar $$\bar x$$ and \overline $$\overline{xyz}$$, and \vec $$\vec x$$ and \overrightarrow $$\overrightarrow{xy}$$ and \overleftrightarrow $$\overleftrightarrow{xy}$$. For dots, as in $$\frac d{dx}x\dot x = \dot x^2 + x\ddot x$$, use \dot and \ddot.

2. Special characters used for MathJax interpreting can be escaped using the \ character: \\\$ $$\$$, \{ $$\{$$, \_ $$\_$$, etc. If you want \ itself, you should use \backslash (symbol) or \setminus (binary operation) for $$\backslash$$, because \\ is for a new line. ta.stackexchange.com/a/29979/676335 This box is free for everyone to use This answer is free for anyone to use 1. A Pythagorean Triple (PT) is a triple $$(a,b,c)$$ of positive integers for which $$a^2+b^2 = c^2$$. Pythagorean triples can be parameterized by positive integers $$s$$ and $$t$$ by $$(a,b,c) = (2st, s^2-t^2, s^2+t^2)$$ 2. An arithmetic progression of squares (APS) is a triple $$(a^2, b^2, c^2)$$ for which $$b^2 - a^2 = c^2 - b^2$$. This is equivalent to $$a^2+c^2 = 2b^2$$. The value of $$b^2 - a^2 = c^2 - b^2$$ is called the common difference. 3. If $$(a,b,c)$$ is a PT, then $$((b-a)^2, c^2, (b+a)^2)$$ is an APS. The common difference is $$2ab$$. 4. Since $$(2st, s^2-t^2, s^2+t^2)$$ is a PT, then $$((s^2-t^2 - 2st)^2, (s^2+t^2)^2, (s^2-t^2+2st)^2)$$ is an APS. The common difference is $$4st(s^2-t^2)$$ We are given the APS $$(205^2, 425^2, 565^2)$$ with a common difference of $$138600$$. The corresponding PT is $$(385, 180, 425)$$ The common difference is$425^2 -

Notice that removing any edge $$e\in T\setminus T'$$, from $$T$$ itself, will result in two connected components: $$V,U$$. Since $$T'$$ is a tree, it is connected. Therefore, there must be some edge $$e'\in T'$$ between the components $$V$$ and $$U$$.

I claim, the three following things:

1. $$e'\notin T$$
2. By removing $$e$$ from $$T$$ and adding instead $$e'$$, we get a tree $$\hat T$$
3. The tree $$\hat T$$ we get is a minimum spanning tree

Lets prove those statements one by one:

1. Assume towards contradiction that $$e'\in T$$. Then, since $$e\notin T'$$ but $$e'\in T'$$ we can conclude that $$e\neq e'$$. Therefore, removing $$e$$ from $$T$$ will still keep $$e'$$ in $$T$$, and thus $$V$$ and $$U$$ will still be connected - contradiction.

2. both $$V$$ and $$U$$ must be connected, since $$T$$ was. Adding the edge $$e'$$ will connect $$V$$ and $$U$$, hence the resulting graph must be connected. Additionally, the number of edges in the resulting graph is $$|T|-1+1=|T|=n-1$$ where $$n$$ is the number of vertices. Thus, the resulting graph must be a tree.

3. This is the trickiest one to prove. Denote by $$V',U'$$ the subgraph of $$T'$$ on the vertices of $$V,U$$. It can be easily shown that $$w(V')=w(V)$$ ($$w(\cdot)$$ is the weight of the subgraph), since otherwise either $$T$$ or $$T'$$ would not be an MST. The same can be applied for $$w(U')=w(U)$$. Now, notice that $$T'$$ is a tree, and thus there is only one single edge between $$V'$$ and $$U'$$ that is also in $$T'$$, and this edge is $$e'$$ (if there were two edges, the graph would contain a cycle). Clearly, $$$${w(V)+w(U)+w(e)=w(T)=w(T')=w(V')+w(U')+w(e')}$$$$ Substitute into this equation the fact that $$w(V)=w(V')$$ and $$w(U)=w(U')$$, and subtruct from both sides of the equation them. We finally get that $$w(e)=w(e')$$. Now, by the definition of $$\hat T$$, we have $$w(\hat T)=w(V)+w(U)+w(e')=w(V)+w(U)+w(e)=w(T)$$. And since $$w(T)$$ is minimal, then also $$w(\hat T)$$ is. Thus $$\hat T$$ is a minimum spanning tree.

$$j=5^{(5^2)}$$ $$j-Z(j!)=j-($$ $$(j-1)-Z((j-1)!)=(j-1)-Z((j-1)!)$$

$$(5^1-5^0) - Z((5-5^0)!) = 4- 0 = 4$$ $$5^1 - Z(5!) = 5- 1 = 4$$

$$5^5 - Z(5^5!) = 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$$ $$(5^5-5)- Z((5^5-5)!) = (5^5-5)- (5^4-1 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$$ $$(5^5-5^1-5^0)- Z((5^5-5^1-5^0)!) = (5^5-5-1)- (5^4-2 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$$

$$(5^{25}-5) - Z((5^{25}-5)!) = (5^{25}-5)- (5^{24}-1 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$$ $$(5^{25}-5^1-5^0) - Z((5^{25}-6)!) = (5^{25}-5 -1 )- (5^{24}-2 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$$ $$(5^{25}+5^2-5^1) - Z((5^{25}+24)!) = (5^{25}+24 )- (5^{24}+4 + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$$ $$(5^{25}+5^2+5^1) - Z((5^{25}+26)!) = (5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$$

2: $$5^1-5^0, 5^1\mid\mid 5^1-5^0$$ First time a number appearing two times: $$5^1-5^0$$ $$(5^1-5^0+4)$$

3: $$5^5-5^1-5^0, 5^5-5^1, 5^5 \mid\mid 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$$ First time a number appearing three times: Two numbers: $$5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 0$$ $$5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 4$$

4: $$5^{25}-5^1-5^0, 5^{25}-5, 5^{25}+5^2+5^1,5^{25}+5^2+5^1\mid\mid(5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$$ First time ??? a number appearing four times: Five numbers: $$5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+0??????$$ a mild speculation Is this a missing one? $$5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+4$$ $$5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+8$$ $$5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+12$$ $$5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+16$$ $$5^{25}- (5^{24} + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+20$$

Some number near $$5^{125}-(5^{125}+ 5^{124}+ 5 ^{123}+ 5^{122} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$$ appears five times as generated by a number near $$5^{125}-(5^{3}+ 5^{2}+ 5 ^{1} + 1)$$

This answer is free for anyone to use.

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Learn
Perceptron
The Perceptron Algorithm

But one question still remains — how do we tweak the weights optimally? We can’t just play around randomly with the weights until the correct combination magically pops up. There needs to be a way to guarantee that the perceptron improves its performance over time.

This is where the Perceptron Algorithm comes in. The math behind why this works is outside the scope of this lesson, so we’ll directly apply the algorithm to optimally tweak the weights and nudge the perceptron towards zero error.

The most important part of the algorithm is the update rule where the weights get updated:

$weight = weight + (error * input)$

We keep on tweaking the weights until all possible labels are correctly predicted by the perceptron. This means that multiple passes might need to be made through the training_set before the Perceptron Algorithm comes to a halt.

In this exercise, you will continue to work on the .training() method. We have made the following changes to this method from the last exercise:

• foundLine = False (a boolean that indicates whether the perceptron has found a line to separate the positive and negative labels)
• while not foundLine: (a while loop that continues to train the perceptron until the line is found)
• total_error = 0 (to count the total error the perceptron makes in each round)
• total_error += abs(error) (to update the total error the perceptron makes in each round)

### Instructions

1.

If the algorithm doesn’t find an error, the perceptron must have correctly predicted the labels for all points.

Outside the for loop (but inside the while loop), change the value of foundLine to True if total_error equals 0.

2.

In order to update the weight for each inputs, create another for loop (inside the existing for loop) that iterates a loop variable i through a range of self.num_inputs.

3.

Inside the second for loop, update each weight self.weights[i] by applying the update rule:

$weight = weight + (error * inputs)$
4.

Great job! Now give it a try for yourself.

Train cool_perceptron using small_training_set.

You can also print out the optimal weights to see for yourself!


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# A homomorphism problem for flows

 Importance: Medium ✭✭
 Author(s): DeVos, Matt
 Subject: Graph Theory » Coloring » » Nowhere-zero flows
 Keywords: homomorphism nowhere-zero flow tension
\begin{conjecture} Let $M,M'$ be abelian groups and let $B \subseteq M$ and $B' \subseteq M'$ satisfy $B=-B$ and $B' = -B'$. If there is a \Def[homomorphism]{graph homomorphism} from $Cayley(M,B)$ to $Cayley(M',B')$, then every graph with a B-flow has a B'-flow. \end{conjecture}
Definition:Let $G$ be a directed graph, Let $M$ be an abelian group, and let $B$ be a subset of $M$ such that $B=-B$. We say that a flow or a tension $\phi:E(G) \rightarrow M$ is a $B$-flow or a $B$-tension if the range is a subset of $B$. If $\phi$ is a $B$-flow ($B$-tension) of $G$ and we reverse the direction of the edge $e$, then we may obtain a new $B$-flow ($B$-tension) by changing $\phi(e)$ to $-\phi(e)$. Thus, the existence of a $B$-flow or $B$-tension does not depend on the orientation, and we say that an undirected graph has a $B$-flow or a $B$-tension if some (and thus every) orientation of it admits such a map. We define the Cayley graph $Cayley(M,B)$ to be the simple graph with vertex set $M$ in which two vertices $u,v$ are joined by an edge if and only if $u-v \in B$.
It is well known that a graph has a $B$-tension if and only if it has a homomorphism to $Cayley(M,B)$. So, if $M,M',B,B'$ are as in the conjecture and there is a homomorphism from $Cayley(M,B)$ to $Cayley(M',B')$, then every graph G with a $B$-tension has a $B'$-tension. This follows from the previous sentence and the fact that the composition of two homomorphisms is another homomorphism. In essence, the above conjecture states that the same equivalence should hold true for flows.
If $H$ and $H^*$ are directed planar dual graphs (each edge of $H^*$ crosses left to right over the corresponding edge of $H$), then a map $\phi:E(H)->M$ is a tension if and only if the dual map $\phi^*:E(H*)->M$ ($\phi^*$ is given by the rule $\phi^*(e^*)=\phi(e)$) is a flow of $H^*$. Thus, planar duality exchanges flows and tensions. For two undirected planar dual graphs, $G$ and $G^*$ we have that G has a $B$-flow if and only if $G^*$ has a $B$-tension. It follows from this duality and the observation from the previous paragraph, that the above conjecture is true for planar graphs.
This conjecture is also known in the special case when $B=M\setminus \{0\}$ and $B'=M'\setminus \{0\}$. In this case, $Cayley(M,B)$ and $Cayley(M',B')$ are the complete graphs on $|M|$ and $|M'|$ vertices respectively, so there is a homomorphism from $Cayley(M,B)$ to $Cayley(M',B')$ if and only if $|M'|$ is greater than or equal to $|M|$. Thus, in this case the conjecture is equivalent to the assertion that every graph with a nowhere-zero $M$-flow also has a nowhere-zero $M'$-flow if $|M'|$ is at least $|M|$. This statement is true by a result of Tutte.


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# Inverse of a modular matrix

I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. I do it according to this website.

I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). Then I have $$\frac{1}{19}×\begin{pmatrix}4 & -5\\-3 & 1\end{pmatrix}.$$ I calculate that$$-5 \equiv 21 \pmod{26},\ -3 \equiv 23 \pmod{26}.$$

No matter what I do I am not able to get the solution they have on the website, which is$$\begin{pmatrix}2 & 17\\5 & 7\end{pmatrix}.$$

Can someone help me with this? What am I doing wrong?

• You multiply by the reciprocal of $-11$ (which is $7$) not divide by it. – Lord Shark the Unknown Mar 11 '18 at 8:46

The determinant is $-11$, as you mentioned.

The Euclidean algorithm will tell you that the inverse is $7$ or $-19$, since $-11 \times 7 = -77 \equiv 1 \mod 26$.

Now the usual $2 \times 2$ inverse is $\begin{pmatrix}4 & -5 \\ -3 & 1\end{pmatrix}$, and this times $7$ is $\begin{pmatrix}28 & -35 \\ -21 & 7\end{pmatrix}$,which simplifies $\mod 26$ to $\begin{pmatrix} 2 & 17 \\ 5 & 7\end{pmatrix}$.

Reiterate procedure, for your help. Given a matrix $A$, which you want to invert $\mod m$ (where the inverse of $A$ exists $\mod m$),

• First, compute the determinant of the matrix,$\det A$. If $\det A$ is coprime to $m$, then you can be sure that $A$ is invertible $\mod m$.

• Find the inverse of $\det A$ modulo $m$. This we denote by $(\det A)^{-1}$ and will be the unique integer between $0$ and $m$ which satisfies $(\det A) \times (\det A)^{-1} \equiv 1 \mod m$.

• Next, compute the matrix of cofactors of $A$, call this $B$. So, this is the matrix which would have been the usual inverse of $A$, without division by the determinant.

• The matrix $(\det A)^{-1} \times B$ is an inverse to $A$ modulo $m$. You can ensure that all the entries of the above matrix are between $0$ and $m$ for completeness, by division by $26$ and taking the consequent remainder of each entry.

The determinant is $-11 \pmod{26}$.

$$(-7)(-11)=77=26\cdot 3-1\equiv -1 \pmod{26}$$

Hence the inverse is suppose to be $7$ rather than $-7$.

$$7 \begin{bmatrix} 4 & -5 \\ -3 & 4\end{bmatrix} \equiv \begin{bmatrix} 2 & 17 \\ 5 & 3\end{bmatrix} \pmod{26}$$

Doing aritmetic modulo $\;26\;$ all the time, we get that

$$\det A=\begin{vmatrix}1&5\\3&4\end{vmatrix}=4-15=-11=15\implies$$

$$\implies A^{-1}=\frac1{15}\begin{pmatrix}4&-5\\-3&1\end{pmatrix}=7\begin{pmatrix}4&21\\23&1\end{pmatrix}=\begin{pmatrix}2&17\\5&7\end{pmatrix}$$


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${\boldsymbol {\boldsymbol c}}$ ${\boldsymbol {\overline{\boldsymbol c}}}$ MESONS(including possibly non- ${\boldsymbol {\boldsymbol q}}$ ${\boldsymbol {\overline{\boldsymbol q}}}$ states) INSPIRE search

# ${{\boldsymbol \psi}{(4260)}}$ $I^G(J^{PC})$ = $0^-(1^{- -})$

also known as ${{\mathit Y}{(4260)}}$; was ${{\mathit X}{(4260)}}$
This state shows properties different from a conventional ${{\mathit q}}{{\overline{\mathit q}}}$ state. A candidate for an exotic structure. See the review on non- ${{\mathit q}}{{\overline{\mathit q}}}$ states. Seen in radiative return from ${{\mathit e}^{+}}{{\mathit e}^{-}}$ collisions at $\sqrt {s }$ = $9.54 - 10.58$ GeV by AUBERT,B 2005I, HE 2006B, and YUAN 2007 , and in ${{\mathit e}^{+}}{{\mathit e}^{-}}$ collisions at $\sqrt {s }$ $\approx{}$ 4.26 GeV by COAN 2006 . Possibly seen by AUBERT 2006 in ${{\mathit B}^{-}}$ $\rightarrow$ ${{\mathit K}^{-}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit J / \psi}}$ . See also the review on "Spectroscopy of mesons containing two heavy quarks."
 ${{\mathit \psi}{(4260)}}$ MASS $4230 \pm8$ MeV (S = 2.9)
 ${{\mathit \psi}{(4260)}}$ WIDTH $55 \pm19$ MeV (S = 4.4)


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# Interpretation of the regression coefficient of a proportion type independent variable

If I want to use some proportion type independent variables in a logistic regression, then what will be the interpretation of the regression coefficients corresponding to those proportion type variables? Will that mean- "The change in log odds for per unit change in the proportions"?

But what will be meant by "per unit change" in this case? As the proportions lie within [0,1], I am getting a little confused with what a "per unit change" will mean in this scale. Does it mean 0.01 or 1%? (I am sorry for my noob thoughts!)

In my data the range of the proportions is 0 to 1, not multiplied by 100. Do I need to multiply them by 100? So that I can say "per unit change" means 1% change? I have seen that the coefficients do differ in scale if I multiply the proportions by 100. For example, a coefficient of -1.3 for proportions becomes -0.013 for percentages (when the proportions are multiplied by 100).

-

The interpretation for the regression coefficient is always for a 1 unit change regardless of what a "unit" is. In your case, if the IV is a proportion falling between 0 and 1, a one unit change is the same as 100%.

If instead you want to look at the "effect" of a 1% change, simply multiply your IV by 100 before using it in the regression.

-
Thank you. You are absolutely right. Actually FMZ is also right. "Per unit change" for a proportion type variable should always mean change by 1 (or 100%) unit in the independent variable. Just that for my example, Y goes down by 1.3 units when X (measured in proportions) increases by 1 unit also implies that, Y goes down by 0.013 (1.3/100=-0.013) units when X (measured in proportions) increases by 0.01 (1/100=0.01) units. That makes it consistent with the sense of proportions. Thank you all. – Blain Waan Jan 11 '13 at 11:16

They are the same, aren't they?

Let's take the first model, coefficient is -1.3 and variable is in original scale (0-1). So if the variable increases by 0.01, say from 0.08 to 0.09, then the log odds are down by 1.3*0.01 = 0.013.

Now the second model, coefficient is -0.013 and variable is multiplied by 100 (0 - 100). So if the variable increases by 1 from 8 to 9 (which is actually from 0.08 to 0.09), then the log odds are down by 0.013*1=0.013.

-
Dear FMZ, thank you for your kind reply, but could you please confirm me when I use proportions as independent variables then "per unit change" in the independent variable means an increase by 0.01. I was guessing it, but as it is a continuous variable just like the other continuous variables (which would generally mean an increase by 1), so I was afraid if a unit change meant '+1' or '+0.01' in this case! Probably very immature question, but I would like you to explain this a little. – Blain Waan Jan 11 '13 at 10:15
Ok I have got it. I would like to thank you too for your kind thoughts. – Blain Waan Jan 11 '13 at 11:21


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Math Help - Commutative Ring

1. Commutative Ring

Let I be any set and let R be the collection of all subsets of I. Define addition and multiplication of subsets A,B ⊆ I as follows:
A+B = (A∪B) ∩ [A∩B(with a line over this)] and AB = A∩B. Show that R is a commutative ring under this addition and multiplication.

I know that to be a commutative ring it has to have the following:
closure, associativity, commutativity, distribution, identity and inverses.

Closure - show well-defined
Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
Comm - a+b = b+a and ab = ba
Dist - a(b+c) = ab+ac and (a+b)c = ac+bc

I am not sure how to write them out using the defined subsets above.

2. So, $A + B = (A\cup B)\cap\overline{(A\cap B)}$. When talking about set operations, this is called the symmetric difference and is commonly denoted by $A\vartriangle B$.

Closure - show well-defined
Do you have difficulties with this?

Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
The associativity of intersection is pretty obvious. For addition, it's a little messy. Hint: prove that $x\in (a+b)+c$ iff x belongs to exactly one of a, b, c or to all three, and similarly for $x\in a+(b+c)$.

Comm - a+b = b+a and ab = ba
Follows directly from the definition and commutativity of union and intersection.

Dist - a(b+c) = ab+ac and (a+b)c = ac+bc
Draw a Venn diagram of a, b, c and then prove that $x\in a(b+c)$ iff $x\in ab+ac$.

3. Any help with identity or inverses?

4. Originally Posted by page929
Any help with identity or inverses?
What is $A\Delta A=~?$

What is $A\Delta \emptyset=~?$

5. Originally Posted by Plato
What is $A\Delta A=~?$

What is $A\Delta \emptyset=~?$
$emptyset$
A complement

6. The best way to view this operation is $A\Delta B=(A\cap\overline{B})\cup(B\cap\overline{A}).$
That is the same as $(A\cup B)\cap (\overline{A\cap B})$.

Now $A\Delta \emptyset = \left( {A \cap \overline \emptyset } \right) \cup \left( {\emptyset \cap \overline A } \right) = A$


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1

# If the temperature $T$ of a solid or liquid of volume $V$ is changed by an amount $\Delta T,$ then the volume will change by the amount $\Delta V=\beta V \Delta T,$...

## Question

###### If the temperature $T$ of a solid or liquid of volume $V$ is changed by an amount $\Delta T,$ then the volume will change by the amount $\Delta V=\beta V \Delta T,$ where $\beta$ is called the coefficient of volume expansion. For moderate changes in temperature $\beta$ is taken as constant. Suppose that a tank truck loads 4000 gallons of ethyl alcohol at a temperature of $35^{\circ} \mathrm{C}$ and delivers its load sometime later at a temperature of $15^{\circ} \mathrm{C} .$ Using $\beta=7.5 \t If the temperature$T$of a solid or liquid of volume$V$is changed by an amount$\Delta T,$then the volume will change by the amount$\Delta V=\beta V \Delta T,$where$\beta$is called the coefficient of volume expansion. For moderate changes in temperature$\beta$is taken as constant. Suppose that a tank truck loads 4000 gallons of ethyl alcohol at a temperature of$35^{\circ} \mathrm{C}$and delivers its load sometime later at a temperature of$15^{\circ} \mathrm{C} .$Using$\beta=7.5 \times 10^{-4} /^{\circ} \mathrm{C}$for ethyl alcohol, find the number of gallons delivered. ## Answers #### Similar Solved Questions 5 answers ##### Use spherical coordinates and integration t0 find the volume of the cap of a sphere of radius R with thickness h: See figure.Use the definition for the average value of a function over a region R Use spherical coordinates and integration t0 find the volume of the cap of a sphere of radius R with thickness h: See figure. Use the definition for the average value of a function over a region R... 5 answers ##### Points) Solve the diffcrential equations. #-"=te' , y(1) = e - 1 (b) (1r + 2y"r)dr (2yr? cos y)dy = 0, ulr_Inrt) , y(1) = #+=yr, 9(1) = 1.9(1) =T.constant rate of L{mnin into large tank that (3 points) brine solution of salt flows at dissolved 0.5 kg of salt. The solution inside initially held 100 L of brine solution in which was ratc If the concentration of the tank is kept well stirred and flows out of the tank at the same the mnass of salt in thc tank after salt in the brine en points) Solve the diffcrential equations. #-"=te' , y(1) = e - 1 (b) (1r + 2y"r)dr (2yr? cos y)dy = 0, ulr_Inrt) , y(1) = #+=yr, 9(1) = 1. 9(1) =T. constant rate of L{mnin into large tank that (3 points) brine solution of salt flows at dissolved 0.5 kg of salt. The solution inside ini... 5 answers ##### Find an equation of the line salisfying the follwing conditions. possible, wrile Ihe equation in slopo-intercept (crm. Horizontal, passing through 8,4)Y=4C. y= -8 0 D. X=4 Find an equation of the line salisfying the follwing conditions. possible, wrile Ihe equation in slopo-intercept (crm. Horizontal, passing through 8,4) Y=4 C. y= -8 0 D. X=4... 5 answers ##### (10 pts) A ball is launched vertically from the ground: While the ball is on its way up, it takes the ball 0.2 seconds to pass window of vertical size of 2 m: The bottom of the window pane is 20 m above the ground. ((Use g 10 m/s2) What is the maximum height the ball reaches? What is the launching speed of the ball at the ground? How long does it take for the ball to complete round trip?Tzm1 Zom U6 (10 pts) A ball is launched vertically from the ground: While the ball is on its way up, it takes the ball 0.2 seconds to pass window of vertical size of 2 m: The bottom of the window pane is 20 m above the ground. ((Use g 10 m/s2) What is the maximum height the ball reaches? What is the launching s... 5 answers ##### Ejemplo 3 : Un granjero tiene 750 pies de alambre y desea cercar un terreno rectangular y entonces dividirlo en rectangulos iguales de tal manera que Ia verja de cada uno de estos rectangulos sea paralela a Ia verja del rectangulo inicial. Cual es Ia mayor area posible? Ejemplo 3 : Un granjero tiene 750 pies de alambre y desea cercar un terreno rectangular y entonces dividirlo en rectangulos iguales de tal manera que Ia verja de cada uno de estos rectangulos sea paralela a Ia verja del rectangulo inicial. Cual es Ia mayor area posible?... 5 answers ##### 4. (a) (8pts) Show that the following equation is exactcos I cos y + 2r) dx (sin T sin ! +29) dy =(1Opts) Find the general solutions of the above equation_ 4. (a) (8pts) Show that the following equation is exact cos I cos y + 2r) dx (sin T sin ! +29) dy = (1Opts) Find the general solutions of the above equation_... 5 answers ##### Find all values of 1 satisfying In[(21 + 1)(r + 2)] = 2In(r + 2). Verify the values of x you found are correct _ Find all values of 1 satisfying In[(21 + 1)(r + 2)] = 2In(r + 2). Verify the values of x you found are correct _... 5 answers ##### Question 19f (r) =22 9x; 9 (#) c + 2. Find (f + 9) (v); (f+ 9) (-5)29 82 | 2; 729x + 2; 728x + 2; 672r? Tx; 85 Question 19 f (r) =22 9x; 9 (#) c + 2. Find (f + 9) (v); (f+ 9) (-5) 29 82 | 2; 72 9x + 2; 72 8x + 2; 67 2r? Tx; 85... 5 answers ##### Two trees are on opposite sides of river; as shown in the figure below_ baseline of 100 ft is measured from tree and from that position the angle & to Tz measured to be 28,98 If the baseline perpendicular to the line segment between T and Tz, find the distance between the two trees (Round your answer t0 one decimal place:)RiverICO ft Two trees are on opposite sides of river; as shown in the figure below_ baseline of 100 ft is measured from tree and from that position the angle & to Tz measured to be 28,98 If the baseline perpendicular to the line segment between T and Tz, find the distance between the two trees (Round your a... 5 answers ##### Find the arca under the standard normal distribution curve between Table E and enter the answcr to decimal places.0 and z =-.UseThe area between the two values Find the arca under the standard normal distribution curve between Table E and enter the answcr to decimal places. 0 and z =-.Use The area between the two values... 5 answers ##### A social service agency plans to conduct a survey t0 determine the mean income of its clients The director of the agency prefers that you measure the mean income very accurately, t0 within +/-5300. From a sample taken 1 year ag0, you estimate that the standard deviation of income for this population is about$4,500. Your job is t0 ligure out the necessary sample size t0 reduce sampling error to +/-5300. What sample size should be drawn to meet the director's requirement at the 94 percent l
A social service agency plans to conduct a survey t0 determine the mean income of its clients The director of the agency prefers that you measure the mean income very accurately, t0 within +/-5300. From a sample taken 1 year ag0, you estimate that the standard deviation of income for this populatio...
##### 1no cucolonThe test statistic of z = 82 is obtained when testing the claim that p > 0.5_ Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed: b. Find the P-value Using a significance level of a = 0.05, should we reject Ho or should we fail to reject Ho? Click here_to_view_page ofthe standard normal distribution _table Clickhere to_view_page 2 of the_standard norma distribution tableThis istestb. P-value =(Round to three decimal places as needed:)Choose the correct c
1no cucolon The test statistic of z = 82 is obtained when testing the claim that p > 0.5_ Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed: b. Find the P-value Using a significance level of a = 0.05, should we reject Ho or should we fail to reject Ho? Click here_to_v...
##### Locate the absolute extrema of the function f(x)= -x^2-4x+3 onthe closed interval (-4,4).
locate the absolute extrema of the function f(x)= -x^2-4x+3 on the closed interval (-4,4)....
##### Use the given volume of a box and its length and width to express the height of the box algebraically.Volume is $10 x^{3}+30 x^{2}-8 x-24,$ length is $2,$ width is $x+3.$
Use the given volume of a box and its length and width to express the height of the box algebraically. Volume is $10 x^{3}+30 x^{2}-8 x-24,$ length is $2,$ width is $x+3.$...
##### NOTESASk Your TEACHERwing are8 of 500 ft? traveling at 5o mi/h experiencer It %f 1700 How much lift woul pln plane withies wings _
NOTES ASk Your TEACHER wing are8 of 500 ft? traveling at 5o mi/h experiencer It %f 1700 How much lift woul pln plane with ies wings _...


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# triangle formula pdf

Details Written by Administrator. A, B and C are vertices and a, b, c are sides of triangle ABC. (i) a+b>c (ii) b+c>a (iii) c+a>b. [��} ,�:$N�����Q�u����)�]p�U��e�S��#S9��T�����ԩLg���kQ�� Ø ‘C’ is the circumcenter of It is equidistant from all three vertices. 1. Acute Triangle: If all the three angles of a triangle are acute i.e., less than 90°, then the triangle is an acute-angled triangle. 3 In Right angle Triangle :- Orthocenter lies on the vertex, where 90° angle is formed. The distance between circumcenter and one of vertices of triangle is called circumradius. Obtuse Triangle: If any one of the three angles of a triangle is obtuse (greater than 90°), then that particular triangle is said to be an obtuse angled triangle. �2�%�;�N� ek� o�� endstream endobj 128 0 obj 786 endobj 74 0 obj << /Type /Page /Parent 69 0 R /Resources 75 0 R /Contents [ 79 0 R 81 0 R 83 0 R 85 0 R 87 0 R 89 0 R 119 0 R 121 0 R ] /MediaBox [ 0 0 595 842 ] /CropBox [ 0 0 595 842 ] /Rotate 0 >> endobj 75 0 obj << /ProcSet [ /PDF /Text ] /Font << /F1 106 0 R /F3 94 0 R /F5 98 0 R /F7 115 0 R /F9 112 0 R /TT2 111 0 R >> /ExtGState << /GS1 123 0 R /GS2 124 0 R >> /ColorSpace << /Cs5 110 0 R /Cs9 76 0 R /Cs10 77 0 R >> >> endobj 76 0 obj [ /Separation /PANTONE#20543#20CVC 110 0 R 126 0 R ] endobj 77 0 obj [ /Separation /PANTONE#20653#20CVC 110 0 R 122 0 R ] endobj 78 0 obj 1356 endobj 79 0 obj << /Filter /FlateDecode /Length 78 0 R >> stream Ø Position of circum center in different triangles. Position of circum center in different triangles. a) sin–1 (–x) = – sin–1x b) cos–1 (–x) = π – cosec–1x c) cosec–1 (–x) = – cosec–1x d) sec–1 (–x) = π – sec–1x e) tan–1 (–x) = π – cot–1x Frame 12-23 Centroids from Parts Consider the scalene triangle below as being the difference of two right triangles. Download the Chapter wise Important Maths Formulas and Equations to Solve the Problems Easily and Score More Marks in Your CBSE Board Exams. Triangle angle calculator is a safe bet if you want to know how to find the angle of a triangle. area of a triangle formula, A=1/2Bh For any triangle ABC with side a opposite A, side b opposite B and side c opposite C, height h is represented by a line perpendicular to the base of the triangle. :- A triangle which has two equal sides is called isosceles triangles. Area of Triangle = ½ × base × height. Where, a, b, c indicates the sides of the triangle. Definition : A closed figure formed by three sides. Syndicate Bank PGDBF PO Interview Details. Quadratic formula I L 2 F U 2 F T1 = T 6 E > T E ? ���yY+�d�� U����/�K{�r�&��-�$��!��)����M�l͑E�sL�p\ r��*ժ޳P-c�eιf%a�i��*����+�^�Y�dNxx�,cY��zb���O���VF������kϲ � .��T��'!�����^q�w�:?e��L�m>��Z�B$8��&o����F]«}:S��7>� �BX���r�1�"0���c|��1T��iS g1~�2� �f�D�;B7��X����a�0W��ް��D�����$�nHbNbf���M~����R��O��uC�$�>�'���)�6:D1,bX�s����"ہ����#�\���}o�ἁ1��� Whether you have three sides of a triangle given, two sides and an angle or just two angles, this tool is a solution to your geometry problems. Copyright © 2016- 2021 Paperadda. Incenter is the center of largest circle. Centroid always forms inside in all types of triangle. The resultant vector is known as the composition of a vector. Locate their centroids, both at one-third the altitude and reason that the centroid of the entire triangle lies one-third the altitude above the base. TRIANGLE. Ø Sum of two sides of a triangle is always greater than third side of that triangle. A triangle which has all three sides equal is called equilateral triangle. To find the area of the triangle on the left, substitute the base and the height into the formula for area. Area of an Isosceles Triangle Worksheets. If a circle is drawn, taking ‘C’ as a centre and ‘R’ as radius. $$Area = \frac{1}{2} (base \cdot height) \\ =\frac{1}{2} (22 \cdot 26.8) \\ = 294.8 \text{ inches squared}$$ Relationship of sides to interior angles in a triangles. ֚w�]X���EOhq�C�cV���B��\y&�u-͋�Tq�;�x��n���ۣO9�^���/o0S�5m�!Gˉ��~I����9E�j׉ۓr�g��UV�{ƥ]-���.<4�V��Ι~3��f�|�tSr���2�휬�+�ǥ%���tV?o�+�*t�{E;疛]���~N%'�j'���8 u���D6�}�d��^Xʝ�t��@}�G� [Shortcut-Count the number of triangles embedded inside of the triangle and count horizontal blocks and put it in the formula given below to know total number of triangles in this type of figures] Number of triangles=4n+m where n= number of triangles embedded inside of the triangle and m= number of horizontal blocks In the figure below, It is a trusted platform for students and government exam aspirants. In obtuse Angle Triangle:- Orthocenter lies outside the triangle. Altitude:- An Altitude is a line which passes through a vertex of triangle and meets the opposite side at right angle. Here, base = a, and height = h. Now, apply Pythagoras Theorem in the triangle. The meaning of “Mensuration” in maths is nothing but it is the act of measurement and it is a branch of mathematics associated with the calculation of parameters like Surface Area, Area and Perimeter formula for all shapes pdf including Volume of different geometrical shapes by using mensuration all formulae. Triangle Area = 1/2 of the base X the height A = bh Perimeter = a + b + c (add the length of the three sides) P = Trapezoid Area = 1/2 of the base X the height A = ()h Perimeter = add lengths of all sides a + b1 + b2 + c Circle Radius = the distance from the center to a point on the circle (r). (3) In Right angle triangle:- It lies on the midpoint of hypotenuse. (2) In obtuse-angle triangle: It lies outside the triangle. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first. The shortest side is always opposite the smallest interior angle 2. Then it will pass through three vertices of triangle. Vector addition is defined as the geometrical sum of two or more vectors as they do not follow regular laws of algebra. In this … 2 Incenter (I): Intersection point of angle bisectors of all interior angles in a triangle is called incentre. Note: the remaining two angles of an obtuse angled triangle are always acute. Sides of a right triangle, formulas . :- A triangle which has three different sides is called scalene triangle. © Corbettmaths 2018 cm² Work out the area of this triangle Work out the area of this triangle cm² Orthocenter lies on the vertex, where 90° angle is formed. :- It lies on the midpoint of hypotenuse. Ø If a circle is drawn, taking ‘C’ as a centre and ‘R’ as radius. If practice in finding the area of an isosceles triangle is what you are looking for, then this is the place to be. H�bfcgc�ed@ A�;� ��ݹ �f?�^�uA� The point where the three attitudes of a triangle intersect is called orthocenter. A triangle which have all angles less than 90° is called acute angle triangle. H�t�ˎ&5����!k$B;���H��@���@� -���9���hĪ�O*߱�]q}���֮�Lf�֑~v����,i�|�o\_�{9���Q{yƯ�����r�N:�1�u�}�h��laeḶ9V_��'���F��3ĝu5>3N��N�M�w(~�1^��Z��a�l�㜶��6��iH|�^Wl�.s������R�>�_���zz�?����. This circle is called circum circle of triangle. Triangle formulae A common mathematical problem is to ﬁnd the angles or lengths of the sides of a triangle when some, but not all of these quantities are known. Triangle law of vector addition is one of the vector addition laws. Centroid divides the median in the ratio 2 : 1. A triangle is defined as basic polygon with three edges and three vertices. The long side is always opposite the largest interior angle. Area of Triangle = ½ × base × height. Ø The distance between circumcenter and one of vertices of triangle is called circumradius. The probability density function of a triangular distribution The formula for the probability density function is {a=1 c=6 b=9 3 Circumcenters:- Intersection point of perpendicular bisectors of three sides of a triangle, is called circumcentre. - Line  joining from vertex to midpoint of opposite side is called median. Triangle formulae mc-TY-triangleformulae-2009-1 A common mathematical problem is to ﬁnd the angles or lengths of the sides of a triangle when some, but not all of these quantities are known. :- Intersection point of perpendicular bisectors of three sides of a triangle, is called circumcentre. If r is the in-radius and R is the circumradius of the triangle ABC, then 2(r + R) equals -[AIEEE-2005] the angle A . %PDF-1.2 %���� Formula For a triangle having sides of length a, b, and c and area K, we have where s is the triangle’s semiperimeter. Vedantu.com is No.1 Online Tutoring Company in India provides Free PDF of Important Math Formula for Class 6 to 12 CBSE Board Prepared by Expert Mathematics Teacher. 72 0 obj << /Linearized 1 /O 74 /H [ 1543 895 ] /L 256252 /E 118893 /N 13 /T 254694 >> endobj xref 72 57 0000000016 00000 n 0000001488 00000 n 0000002438 00000 n 0000002647 00000 n 0000002881 00000 n 0000002952 00000 n 0000003023 00000 n 0000003045 00000 n 0000004479 00000 n 0000004501 00000 n 0000005767 00000 n 0000005789 00000 n 0000007017 00000 n 0000007039 00000 n 0000008249 00000 n 0000008271 00000 n 0000009615 00000 n 0000009637 00000 n 0000010961 00000 n 0000011180 00000 n 0000026025 00000 n 0000026208 00000 n 0000026617 00000 n 0000026824 00000 n 0000027421 00000 n 0000043330 00000 n 0000043533 00000 n 0000043837 00000 n 0000045253 00000 n 0000045466 00000 n 0000060943 00000 n 0000061312 00000 n 0000062215 00000 n 0000079604 00000 n 0000079724 00000 n 0000080179 00000 n 0000080393 00000 n 0000080802 00000 n 0000081038 00000 n 0000081219 00000 n 0000081649 00000 n 0000081847 00000 n 0000081928 00000 n 0000082520 00000 n 0000082818 00000 n 0000098218 00000 n 0000098440 00000 n 0000098463 00000 n 0000099747 00000 n 0000099770 00000 n 0000100962 00000 n 0000102079 00000 n 0000102158 00000 n 0000102236 00000 n 0000118100 00000 n 0000001543 00000 n 0000002416 00000 n trailer << /Size 129 /Info 68 0 R /Root 73 0 R /Prev 254684 /ID[] >> startxref 0 %%EOF 73 0 obj << /Type /Catalog /Pages 70 0 R >> endobj 127 0 obj << /S 904 /Filter /FlateDecode /Length 128 0 R >> stream Centroid always forms inside in all types of triangle. Difference of two sides of a triangle is always smaller than the third side of that triangle. Divide the triangle into two right triangles. a 2 = h 2 + (a/2) 2. Plug in the integer, or decimal dimensions in the area of a triangle formula A = 1/2 * b * h and solve for the area. This circle is called circum circle of triangle. That will fit inside the triangle and touch all three sides. The peak is at c=6 with a function value of 0.25. ⇒ h 2 = a 2 – (a 2 /4) ⇒ h 2 = (3a 2 )/4. If SAS is given and h is unknown, m A can be written sin A = h/b Therefore, Multiplying produces b sin A = h Substitute into the formula: A = ½ c (b sinA) 2. Here, we will discuss various triangles with triangle formula. The graph below shows the probability density function of a triangle distribution with a=1, b=9 and c=6. opposite sin hypotenuse q= hypotenuse csc opposite q= adjacent cos hypotenuse q= hypotenuse sec adjacent q= opposite tan adjacent q= adjacent cot opposite q= Unit circle definition For this definition q is any angle. :- A triangle in which one of angles is 90° is called right angle triangle. It is also useful to be able to calculate the area of a triangle from some of this information. Password Protect PDF Password Protect PDF; Ringtone Download. Distance Formula - Triangles Sheet 1 Score : Printable Math Worksheets @ www.mathworksheets4kids.com Name : 2) 3) 4) 1) Show that the points A(7, 5), B(2, 3) and C(6, ±7) are the vertices of a right triangle. 2020In any equilateral , three circles of radii one are touching to the sides given as in the figure then area of the [IIT-2005] Paperadda.com is an online education and examination portal. 2. Drawing a line parallel to BC through point A. : The exterior angle of a triangle is equal sum of two opposite interior angles. All rights reserved. Triangles are also divided into different types based on the measurement of its sides and angles. The longest side is always opposite the largest interior angle Ø AD, BE and CF are altitudes and ‘O’ is orthocenter of, Ø Position of Orthocenter in different triangle:-. (The letter K is used for the area of the triangle to avoid confusion when using the letter A to name an angle of a triangle.) K = s(s−a)(s −b)(s −c) Heron’s Proof: Part A Let ABC be an arbitrary triangle such that side AB is at least as long as the other two sides. Or, h = ½ (√3a) Now, put the value of “h” in the area of the triangle equation. 1. Right triangle definition For this definition we assume that 0 2 p < c. b+c> a. c+a> b. In Acute Angle triangle : Orthocenter lies inside the triangle. Three additional categories of area formulas are useful. To find the area of an equilateral triangle, we can use the Pythagorean Theorem to get the height of the triangle and then use formula $$A = \frac{1}{2}bh$$ or we can use the following formula: The formula for the area of an equilateral triangle (with all sides congruent) is equal to $$A = \frac{{{s^2}\sqrt 3 }}{4}$$ The length of the sides, as well as all three angles, will have different values. In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Then it will pass through three vertices of triangle. Ø A, B and C are vertices and a, b, c are sides of triangle ABC. [IIT-1993] (A) /3 (B) (C) /2 (D) Q. 1. Find the height of the triangle using the Pythagorean theorem. Hence, the formula for the Perimeter of a Triangle when all sides are given is, P= a+b+c. ‘C’ is the circumcenter of It is equidistant from all three vertices. The most common formula for finding the area of a triangle is K = ½ bh, where K is the area of the triangle, b is the base of the triangle, and h is the height. It is also useful to be able to calculate the area of a triangle from some of this information. - A triangle in which one of angles is obtuse angle  is called obtuse angle triangle. 52 Special Triangles (45⁰‐45⁰‐90⁰ Triangle, 30⁰‐60⁰‐90⁰ Triangle) 53 Trigonometric Functions and Special Angles 54 Trigonometric Function Values in Quadrants II, III, and IV 55 Graphs of Trigonometric Functions 56 Vectors 57 Operating with Vectors Version 3.2 Page 3 of 82 August 28, 2018 Median divides area of triangle in two equal points. An Altitude is a line which passes through a vertex of triangle and meets the opposite side at right angle. AD, BE and CF are altitudes and ‘O’ is orthocenter of, Position of Orthocenter in different triangle:-. : sum of all interior angle of a triangle is 180°. The shortest side is always opposite the smallest interior angle. L0 : T F D ; 6 E : U F G ; 6 L N 6 where (x1,y1) and (x2,y2) are two points on a coordinate plane Where a and b are coefficients and c is constant Where r is the radius and (h, k) is the center Where a and b are coefficients and c is constant Sum of two sides of a triangle is always greater than third side of that triangle.

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Surface tension has dimensions of

It has a dimension of force per unit length.

Surface tension is given as:

 $$Surface\;tension=\frac{Force}{Length}$$

Dimension of force = M1L1T-2
Dimension of length = L1
Therefore, dimension of surface tension = M1L0T-2


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# Write the following sequence as a recurrence relation

Write the following sequence as a recurrence relation (with sufficient initial values specified):

$$b_n=1-\frac{1}{2^n} \forall n\in\mathbb{N}^*$$

I think I am suppose to use induction (e.g. n-1) to prove the recurrence, but I am confused, how?

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There are infinite ways to create a recurrence relation. –  Aryabhata Feb 16 '11 at 22:42
This is just my homework I copied straight out of the book...I believe it wants the form asubn = asubn-1 + ???. (stating asub1 = ???). –  Mr_CryptoPrime Feb 16 '11 at 22:46
So what is $b_n - b_{n-1}$? What is $b_1$ or $b_0$? –  Henry Feb 17 '11 at 0:04

HINT $\$ Eliminate $\rm\ 1/2^n\$ from the equations for $\rm\ b_n,\ b_{n+1}\$
If you wrote $$\frac{1}{2^{n}} = 1 - b_{n}$$ and similarly for the $n-1$ case then you could probably see a relationship between $b_{n-1}$ and $b_n$