text:
# Properties

 Label 9.40.a.c Level $9$ Weight $40$ Character orbit 9.a Self dual yes Analytic conductor $86.706$ Analytic rank $1$ Dimension $3$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$9 = 3^{2}$$ Weight: $$k$$ $$=$$ $$40$$ Character orbit: $$[\chi]$$ $$=$$ 9.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$86.7055962508$$ Analytic rank: $$1$$ Dimension: $$3$$ Coefficient field: $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ Defining polynomial: $$x^{3} - 3876249523 x - 18467420411022$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$2^{8}\cdot 3^{6}\cdot 5\cdot 7\cdot 13$$ Twist minimal: no (minimal twist has level 3) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( -177858 + \beta_{1} ) q^{2} + ( 319147551244 - 227081 \beta_{1} + \beta_{2} ) q^{4} + ( 17793991314810 - 12092638 \beta_{1} - 54 \beta_{2} ) q^{5} + ( -525770892074176 - 12596523938 \beta_{1} + 56758 \beta_{2} ) q^{7} + ( -149112778455452232 + 182657330678 \beta_{1} - 533574 \beta_{2} ) q^{8} +O(q^{10})$$ $$q +(-177858 + \beta_{1}) q^{2} +(319147551244 - 227081 \beta_{1} + \beta_{2}) q^{4} +(17793991314810 - 12092638 \beta_{1} - 54 \beta_{2}) q^{5} +(-525770892074176 - 12596523938 \beta_{1} + 56758 \beta_{2}) q^{7} +(-149112778455452232 + 182657330678 \beta_{1} - 533574 \beta_{2}) q^{8} +(-13289593742193286260 - 3323523968902 \beta_{1} + 4457984 \beta_{2}) q^{10} +($$$$17\!\cdots\!80$$$$- 2617390166972 \beta_{1} + 42093972 \beta_{2}) q^{11} +($$$$18\!\cdots\!30$$$$- 3296990601565580 \beta_{1} + 2302171012 \beta_{2}) q^{13} +(-$$$$10\!\cdots\!84$$$$+ 22915977673056320 \beta_{1} - 29992453632 \beta_{2}) q^{14} +($$$$40\!\cdots\!12$$$$- 247808314916500164 \beta_{1} - 203561787228 \beta_{2}) q^{16} +(-$$$$24\!\cdots\!86$$$$+ 1432579830459091908 \beta_{1} - 73008839916 \beta_{2}) q^{17} +(-$$$$35\!\cdots\!08$$$$- 3709293048070314324 \beta_{1} + 14019280007196 \beta_{2}) q^{19} +(-$$$$10\!\cdots\!20$$$$- 4685500071252208714 \beta_{1} + 24996949090938 \beta_{2}) q^{20} +(-$$$$33\!\cdots\!08$$$$+$$$$19\!\cdots\!84$$$$\beta_{1} - 15518897927168 \beta_{2}) q^{22} +(-$$$$13\!\cdots\!88$$$$- 71070379474158455740 \beta_{1} - 864912659360364 \beta_{2}) q^{23} +(-$$$$39\!\cdots\!25$$$$- 37236871224450578840 \beta_{1} - 3116005674041720 \beta_{2}) q^{25} +(-$$$$30\!\cdots\!92$$$$+$$$$29\!\cdots\!78$$$$\beta_{1} - 4002589901546496 \beta_{2}) q^{26} +($$$$21\!\cdots\!52$$$$-$$$$16\!\cdots\!88$$$$\beta_{1} + 905414279433792 \beta_{2}) q^{28} +(-$$$$24\!\cdots\!54$$$$+$$$$60\!\cdots\!54$$$$\beta_{1} - 10883586973384674 \beta_{2}) q^{29} +(-$$$$12\!\cdots\!52$$$$+$$$$13\!\cdots\!26$$$$\beta_{1} - 51025473408711938 \beta_{2}) q^{31} +(-$$$$12\!\cdots\!04$$$$-$$$$16\!\cdots\!32$$$$\beta_{1} + 107917356575846952 \beta_{2}) q^{32} +($$$$12\!\cdots\!48$$$$-$$$$34\!\cdots\!14$$$$\beta_{1} + 1454956528831466496 \beta_{2}) q^{34} +(-$$$$91\!\cdots\!40$$$$-$$$$42\!\cdots\!08$$$$\beta_{1} + 2423830555879147836 \beta_{2}) q^{35} +($$$$96\!\cdots\!74$$$$-$$$$33\!\cdots\!92$$$$\beta_{1} - 3950416425949998144 \beta_{2}) q^{37} +(-$$$$24\!\cdots\!64$$$$+$$$$22\!\cdots\!08$$$$\beta_{1} - 8006104235315837952 \beta_{2}) q^{38} +($$$$51\!\cdots\!00$$$$+$$$$19\!\cdots\!20$$$$\beta_{1} - 14797692611200750940 \beta_{2}) q^{40} +($$$$46\!\cdots\!42$$$$-$$$$18\!\cdots\!28$$$$\beta_{1} - 41049028078685916228 \beta_{2}) q^{41} +(-$$$$80\!\cdots\!76$$$$-$$$$20\!\cdots\!56$$$$\beta_{1} + 42956960985072135700 \beta_{2}) q^{43} +($$$$71\!\cdots\!04$$$$-$$$$48\!\cdots\!16$$$$\beta_{1} +$$$$17\!\cdots\!72$$$$\beta_{2}) q^{44} +(-$$$$34\!\cdots\!52$$$$-$$$$48\!\cdots\!44$$$$\beta_{1} +$$$$19\!\cdots\!12$$$$\beta_{2}) q^{46} +($$$$36\!\cdots\!68$$$$+$$$$27\!\cdots\!32$$$$\beta_{1} -$$$$15\!\cdots\!28$$$$\beta_{2}) q^{47} +($$$$30\!\cdots\!81$$$$-$$$$70\!\cdots\!80$$$$\beta_{1} -$$$$10\!\cdots\!68$$$$\beta_{2}) q^{49} +($$$$39\!\cdots\!50$$$$-$$$$16\!\cdots\!85$$$$\beta_{1} +$$$$91\!\cdots\!20$$$$\beta_{2}) q^{50} +($$$$19\!\cdots\!96$$$$-$$$$30\!\cdots\!10$$$$\beta_{1} +$$$$28\!\cdots\!50$$$$\beta_{2}) q^{52} +(-$$$$20\!\cdots\!78$$$$-$$$$33\!\cdots\!18$$$$\beta_{1} -$$$$29\!\cdots\!98$$$$\beta_{2}) q^{53} +($$$$24\!\cdots\!80$$$$-$$$$22\!\cdots\!24$$$$\beta_{1} -$$$$78\!\cdots\!92$$$$\beta_{2}) q^{55} +(-$$$$12\!\cdots\!00$$$$+$$$$99\!\cdots\!44$$$$\beta_{1} -$$$$50\!\cdots\!28$$$$\beta_{2}) q^{56} +($$$$93\!\cdots\!28$$$$-$$$$28\!\cdots\!62$$$$\beta_{1} +$$$$93\!\cdots\!36$$$$\beta_{2}) q^{58} +($$$$25\!\cdots\!52$$$$-$$$$10\!\cdots\!28$$$$\beta_{1} -$$$$27\!\cdots\!56$$$$\beta_{2}) q^{59} +(-$$$$23\!\cdots\!66$$$$+$$$$74\!\cdots\!00$$$$\beta_{1} +$$$$14\!\cdots\!48$$$$\beta_{2}) q^{61} +($$$$11\!\cdots\!72$$$$-$$$$28\!\cdots\!92$$$$\beta_{1} +$$$$14\!\cdots\!60$$$$\beta_{2}) q^{62} +(-$$$$11\!\cdots\!52$$$$+$$$$61\!\cdots\!32$$$$\beta_{1} -$$$$87\!\cdots\!04$$$$\beta_{2}) q^{64} +($$$$23\!\cdots\!20$$$$-$$$$87\!\cdots\!96$$$$\beta_{1} +$$$$12\!\cdots\!32$$$$\beta_{2}) q^{65} +(-$$$$17\!\cdots\!24$$$$+$$$$39\!\cdots\!84$$$$\beta_{1} +$$$$35\!\cdots\!32$$$$\beta_{2}) q^{67} +(-$$$$37\!\cdots\!64$$$$+$$$$10\!\cdots\!30$$$$\beta_{1} -$$$$74\!\cdots\!34$$$$\beta_{2}) q^{68} +($$$$12\!\cdots\!40$$$$+$$$$63\!\cdots\!68$$$$\beta_{1} -$$$$78\!\cdots\!56$$$$\beta_{2}) q^{70} +(-$$$$28\!\cdots\!92$$$$-$$$$73\!\cdots\!04$$$$\beta_{1} +$$$$74\!\cdots\!88$$$$\beta_{2}) q^{71} +($$$$21\!\cdots\!38$$$$+$$$$25\!\cdots\!92$$$$\beta_{1} +$$$$69\!\cdots\!00$$$$\beta_{2}) q^{73} +(-$$$$29\!\cdots\!04$$$$-$$$$45\!\cdots\!06$$$$\beta_{1} -$$$$21\!\cdots\!00$$$$\beta_{2}) q^{74} +($$$$42\!\cdots\!12$$$$-$$$$37\!\cdots\!04$$$$\beta_{1} -$$$$29\!\cdots\!04$$$$\beta_{2}) q^{76} +($$$$73\!\cdots\!12$$$$-$$$$26\!\cdots\!32$$$$\beta_{1} +$$$$91\!\cdots\!40$$$$\beta_{2}) q^{77} +(-$$$$54\!\cdots\!00$$$$-$$$$11\!\cdots\!74$$$$\beta_{1} +$$$$11\!\cdots\!38$$$$\beta_{2}) q^{79} +($$$$62\!\cdots\!60$$$$+$$$$17\!\cdots\!12$$$$\beta_{1} -$$$$72\!\cdots\!04$$$$\beta_{2}) q^{80} +(-$$$$16\!\cdots\!12$$$$-$$$$10\!\cdots\!66$$$$\beta_{1} -$$$$59\!\cdots\!24$$$$\beta_{2}) q^{82} +(-$$$$19\!\cdots\!16$$$$-$$$$12\!\cdots\!88$$$$\beta_{1} +$$$$40\!\cdots\!00$$$$\beta_{2}) q^{83} +(-$$$$17\!\cdots\!60$$$$-$$$$47\!\cdots\!32$$$$\beta_{1} +$$$$26\!\cdots\!44$$$$\beta_{2}) q^{85} +(-$$$$28\!\cdots\!00$$$$-$$$$62\!\cdots\!88$$$$\beta_{1} -$$$$33\!\cdots\!56$$$$\beta_{2}) q^{86} +(-$$$$34\!\cdots\!88$$$$+$$$$37\!\cdots\!28$$$$\beta_{1} -$$$$93\!\cdots\!28$$$$\beta_{2}) q^{88} +(-$$$$60\!\cdots\!62$$$$+$$$$73\!\cdots\!36$$$$\beta_{1} -$$$$13\!\cdots\!40$$$$\beta_{2}) q^{89} +($$$$77\!\cdots\!08$$$$-$$$$10\!\cdots\!60$$$$\beta_{1} +$$$$11\!\cdots\!64$$$$\beta_{2}) q^{91} +(-$$$$32\!\cdots\!44$$$$+$$$$10\!\cdots\!88$$$$\beta_{1} -$$$$67\!\cdots\!28$$$$\beta_{2}) q^{92} +(-$$$$41\!\cdots\!40$$$$+$$$$29\!\cdots\!80$$$$\beta_{1} +$$$$75\!\cdots\!36$$$$\beta_{2}) q^{94} +(-$$$$28\!\cdots\!00$$$$+$$$$34\!\cdots\!60$$$$\beta_{1} +$$$$78\!\cdots\!80$$$$\beta_{2}) q^{95} +(-$$$$33\!\cdots\!14$$$$+$$$$51\!\cdots\!24$$$$\beta_{1} +$$$$22\!\cdots\!12$$$$\beta_{2}) q^{97} +(-$$$$64\!\cdots\!70$$$$-$$$$64\!\cdots\!91$$$$\beta_{1} -$$$$39\!\cdots\!56$$$$\beta_{2}) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$3q - 533574q^{2} + 957442653732q^{4} + 53381973944430q^{5} - 1577312676222528q^{7} - 447338335366356696q^{8} + O(q^{10})$$ $$3q - 533574q^{2} + 957442653732q^{4} + 53381973944430q^{5} - 1577312676222528q^{7} - 447338335366356696q^{8} - 39868781226579858780q^{10} +$$$$53\!\cdots\!40$$$$q^{11} +$$$$54\!\cdots\!90$$$$q^{13} -$$$$31\!\cdots\!52$$$$q^{14} +$$$$12\!\cdots\!36$$$$q^{16} -$$$$72\!\cdots\!58$$$$q^{17} -$$$$10\!\cdots\!24$$$$q^{19} -$$$$30\!\cdots\!60$$$$q^{20} -$$$$10\!\cdots\!24$$$$q^{22} -$$$$41\!\cdots\!64$$$$q^{23} -$$$$11\!\cdots\!75$$$$q^{25} -$$$$92\!\cdots\!76$$$$q^{26} +$$$$64\!\cdots\!56$$$$q^{28} -$$$$72\!\cdots\!62$$$$q^{29} -$$$$38\!\cdots\!56$$$$q^{31} -$$$$37\!\cdots\!12$$$$q^{32} +$$$$37\!\cdots\!44$$$$q^{34} -$$$$27\!\cdots\!20$$$$q^{35} +$$$$29\!\cdots\!22$$$$q^{37} -$$$$74\!\cdots\!92$$$$q^{38} +$$$$15\!\cdots\!00$$$$q^{40} +$$$$13\!\cdots\!26$$$$q^{41} -$$$$24\!\cdots\!28$$$$q^{43} +$$$$21\!\cdots\!12$$$$q^{44} -$$$$10\!\cdots\!56$$$$q^{46} +$$$$10\!\cdots\!04$$$$q^{47} +$$$$92\!\cdots\!43$$$$q^{49} +$$$$11\!\cdots\!50$$$$q^{50} +$$$$59\!\cdots\!88$$$$q^{52} -$$$$62\!\cdots\!34$$$$q^{53} +$$$$72\!\cdots\!40$$$$q^{55} -$$$$36\!\cdots\!00$$$$q^{56} +$$$$28\!\cdots\!84$$$$q^{58} +$$$$75\!\cdots\!56$$$$q^{59} -$$$$71\!\cdots\!98$$$$q^{61} +$$$$33\!\cdots\!16$$$$q^{62} -$$$$35\!\cdots\!56$$$$q^{64} +$$$$71\!\cdots\!60$$$$q^{65} -$$$$51\!\cdots\!72$$$$q^{67} -$$$$11\!\cdots\!92$$$$q^{68} +$$$$38\!\cdots\!20$$$$q^{70} -$$$$84\!\cdots\!76$$$$q^{71} +$$$$63\!\cdots\!14$$$$q^{73} -$$$$89\!\cdots\!12$$$$q^{74} +$$$$12\!\cdots\!36$$$$q^{76} +$$$$22\!\cdots\!36$$$$q^{77} -$$$$16\!\cdots\!00$$$$q^{79} +$$$$18\!\cdots\!80$$$$q^{80} -$$$$48\!\cdots\!36$$$$q^{82} -$$$$59\!\cdots\!48$$$$q^{83} -$$$$52\!\cdots\!80$$$$q^{85} -$$$$85\!\cdots\!00$$$$q^{86} -$$$$10\!\cdots\!64$$$$q^{88} -$$$$18\!\cdots\!86$$$$q^{89} +$$$$23\!\cdots\!24$$$$q^{91} -$$$$96\!\cdots\!32$$$$q^{92} -$$$$12\!\cdots\!20$$$$q^{94} -$$$$84\!\cdots\!00$$$$q^{95} -$$$$99\!\cdots\!42$$$$q^{97} -$$$$19\!\cdots\!10$$$$q^{98} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{3} - 3876249523 x - 18467420411022$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$18 \nu$$ $$\beta_{2}$$ $$=$$ $$324 \nu^{2} - 2315430 \nu - 837269896968$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$$$/18$$ $$\nu^{2}$$ $$=$$ $$($$$$\beta_{2} + 128635 \beta_{1} + 837269896968$$$$)/324$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 −59724.7 −4792.65 64517.4
−1.25290e6 0 1.02001e12 6.13018e12 0 3.89397e16 −5.89182e17 0 −7.68052e18
1.2 −264126. 0 −4.79993e11 6.30487e13 0 −4.59086e16 2.71983e17 0 −1.66528e19
1.3 983454. 0 4.17427e11 −1.57969e13 0 5.39162e15 −1.30140e17 0 −1.55355e19
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$3$$ $$-1$$

## Inner twists

This newform does not admit any (nontrivial) inner twists.

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 9.40.a.c 3
3.b odd 2 1 3.40.a.b 3

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
3.40.a.b 3 3.b odd 2 1
9.40.a.c 3 1.a even 1 1 trivial

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{3} + 533574 T_{2}^{2} -$$$$11\!\cdots\!60$$$$T_{2} -$$$$32\!\cdots\!08$$ acting on $$S_{40}^{\mathrm{new}}(\Gamma_0(9))$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$-325448454458769408 - 1161004440960 T + 533574 T^{2} + T^{3}$$
$3$ $$T^{3}$$
$5$ $$61\!\cdots\!00$$$$-$$$$70\!\cdots\!00$$$$T - 53381973944430 T^{2} + T^{3}$$
$7$ $$96\!\cdots\!40$$$$-$$$$18\!\cdots\!44$$$$T + 1577312676222528 T^{2} + T^{3}$$
$11$ $$-$$$$53\!\cdots\!36$$$$+$$$$93\!\cdots\!48$$$$T -$$$$53\!\cdots\!40$$$$T^{2} + T^{3}$$
$13$ $$16\!\cdots\!36$$$$-$$$$63\!\cdots\!92$$$$T -$$$$54\!\cdots\!90$$$$T^{2} + T^{3}$$
$17$ $$-$$$$77\!\cdots\!44$$$$-$$$$24\!\cdots\!12$$$$T +$$$$72\!\cdots\!58$$$$T^{2} + T^{3}$$
$19$ $$-$$$$28\!\cdots\!00$$$$-$$$$78\!\cdots\!40$$$$T +$$$$10\!\cdots\!24$$$$T^{2} + T^{3}$$
$23$ $$-$$$$14\!\cdots\!40$$$$-$$$$32\!\cdots\!96$$$$T +$$$$41\!\cdots\!64$$$$T^{2} + T^{3}$$
$29$ $$12\!\cdots\!00$$$$+$$$$16\!\cdots\!80$$$$T +$$$$72\!\cdots\!62$$$$T^{2} + T^{3}$$
$31$ $$65\!\cdots\!00$$$$-$$$$23\!\cdots\!16$$$$T +$$$$38\!\cdots\!56$$$$T^{2} + T^{3}$$
$37$ $$53\!\cdots\!80$$$$-$$$$19\!\cdots\!84$$$$T -$$$$29\!\cdots\!22$$$$T^{2} + T^{3}$$
$41$ $$72\!\cdots\!00$$$$-$$$$12\!\cdots\!56$$$$T -$$$$13\!\cdots\!26$$$$T^{2} + T^{3}$$
$43$ $$39\!\cdots\!68$$$$+$$$$17\!\cdots\!56$$$$T +$$$$24\!\cdots\!28$$$$T^{2} + T^{3}$$
$47$ $$-$$$$42\!\cdots\!16$$$$+$$$$38\!\cdots\!04$$$$T -$$$$10\!\cdots\!04$$$$T^{2} + T^{3}$$
$53$ $$61\!\cdots\!80$$$$-$$$$51\!\cdots\!96$$$$T +$$$$62\!\cdots\!34$$$$T^{2} + T^{3}$$
$59$ $$56\!\cdots\!00$$$$-$$$$50\!\cdots\!80$$$$T -$$$$75\!\cdots\!56$$$$T^{2} + T^{3}$$
$61$ $$-$$$$27\!\cdots\!52$$$$-$$$$53\!\cdots\!24$$$$T +$$$$71\!\cdots\!98$$$$T^{2} + T^{3}$$
$67$ $$-$$$$88\!\cdots\!76$$$$-$$$$16\!\cdots\!72$$$$T +$$$$51\!\cdots\!72$$$$T^{2} + T^{3}$$
$71$ $$-$$$$61\!\cdots\!12$$$$-$$$$72\!\cdots\!08$$$$T +$$$$84\!\cdots\!76$$$$T^{2} + T^{3}$$
$73$ $$-$$$$42\!\cdots\!60$$$$-$$$$81\!\cdots\!96$$$$T -$$$$63\!\cdots\!14$$$$T^{2} + T^{3}$$
$79$ $$-$$$$23\!\cdots\!00$$$$-$$$$13\!\cdots\!00$$$$T +$$$$16\!\cdots\!00$$$$T^{2} + T^{3}$$
$83$ $$76\!\cdots\!76$$$$+$$$$11\!\cdots\!80$$$$T +$$$$59\!\cdots\!48$$$$T^{2} + T^{3}$$
$89$ $$-$$$$15\!\cdots\!00$$$$+$$$$41\!\cdots\!20$$$$T +$$$$18\!\cdots\!86$$$$T^{2} + T^{3}$$
$97$ $$-$$$$15\!\cdots\!56$$$$-$$$$29\!\cdots\!12$$$$T +$$$$99\!\cdots\!42$$$$T^{2} + T^{3}$$


text:
# If H is a p-group, the order of any H-orbit is a power of p.

This comes from the proof of the third Sylow Theorem in Michael Artin's "Algebra".

Let S be the set of Sylow p-groups in a given group G of order $p^em$. Let H be any Sylow group. If we decompose S into orbits for the operation of conjugation by H, then to establish s=1 modulo p, we only must show no element of S is fixed, since if H is a p-group, the order of any H-orbit is a power of p.

I'm having trouble with this last claim, I can't see why it is true. I don't know how much superfluous information I gave because I really don't understand the inference. Can anyone help?

-

For any group $\,G\,$ acting on any set $\,X\,$ , we have that for any $\,x\in X\,$ :

$$|\mathcal Orb(x)|=[G:G_x]\;,\;\;G_x:=\{g\in G\;;\;gx=x\}$$

Since the index of any subgroup of $\,G\,$ divides the order of $\,G\,$, the claim follows.

-
Aren't we forgetting something? You also have to show that $H$ is the only element of the set $S$ that is fixed, in other words the orbit of $H$ has size 1 and the rest has an orbit size a power of $p$ (as pointed out in the previous answers!) strictly greater than 1. This goes as follows. $H$ fixes an element $K \in S$ iff $H \subseteq N_G(K)$ (the normalizer of $K$ in $G$). Assume the latter. $K$ is a Sylow $p$-subgroup and certainly $K$ is normal in $N_G(K)$, hence its only Sylow $p$-subgroup. Since $H$ is a $p$-subgroup of $N_G(K)$, according to a more general lemma, it must lie in some Sylow $p$-subgroup of $N_G(K)$ for which there is only one choice, $K$. Hence $H \subseteq K$ and this implies $H=K$ because they have the same order. To go short, if $H \neq K$, then the orbit of $K$ is a non-trivial power of $p$. You can now conclude that $|S| \equiv 1$ mod $p$, where the $1$ comes from the orbit of $H$ and all the other orbit sizes are divisible by $p$.
By the way, it is not too hard to show that the orbit size of $K$ actually equals $[H:H\cap K]$ ($= [K:H\cap K]$) –  Nicky Hekster Apr 25 '13 at 15:08

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# Distance geometry

Distance geometry is the characterization and study of sets of points based only on given values of the distances between member pairs.[1][2][3] More abstractly, it is the study of semimetric spaces and the isometric transformations between them. In this view, it can be considered as a subject within general topology.[4]

Historically, the first result in distance geometry is Heron's formula in 1st century AD. The modern theory began in 19th century with work by Arthur Cayley, followed by more extensive developments in the 20th century by Karl Menger and others.

Distance geometry problems arise whenever one needs to infer the shape of a configuration of points (relative positions) from the distances between them, such as in biology,[4] sensor network,[5] surveying, navigation, cartography, and physics.

## Introduction and definitions

The concepts of distance geometry will first be explained by describing two particular problems.

Consider three ground radio stations A, B, C, whose locations are known. A radio receiver is at an unknown location. The times it takes for a radio signal to travel from the stations to the receiver, ${\displaystyle {\displaystyle t_{A},t_{B},t_{C}}}$ , are unknown, but the time differences, ${\displaystyle {\displaystyle t_{A}-t_{B}}}$  and ${\displaystyle {\displaystyle t_{A}-t_{C}}}$ , are known. From them, one knows the distance differences ${\displaystyle c({\displaystyle t_{A}-t_{B}})}$  and ${\displaystyle c({\displaystyle t_{A}-t_{C}})}$ , from which the position of the receiver can be found.

### Second problem: dimension reduction

In data analysis, one is often given a list of data represented as vectors ${\displaystyle \mathbf {v} =(x_{1},\cdots ,x_{n})\in \mathbb {R} ^{n}}$ , and one needs to find out whether they lie within a low-dimensional affine subspace. A low-dimensional representation of data has many advantages, such as saving storage space, computation time, and giving better insight into data.

### Definitions

Now we formalize some definitions that naturally arise from considering our problems.

#### Semimetric space

Given a list of points on ${\displaystyle R=\{P_{0},\cdots ,P_{n}\}}$ , ${\displaystyle n\geq 0}$ , we can arbitrarily specify the distances between pairs of points by a list of ${\displaystyle d_{ij}>0}$ , ${\displaystyle 0\leq i . This defines a semimetric space: a metric space without triangle inequality.

Explicitly, we define a semimetric space as a nonempty set ${\displaystyle R}$  equipped with a semimetric ${\displaystyle d:R\times R\to [0,\infty )}$  such that, for all ${\displaystyle x,y\in R}$ ,

1. Positivity: ${\displaystyle d(x,y)=0}$    if and only if  ${\displaystyle x=y}$ .
2. Symmetry: ${\displaystyle d(x,y)=d(y,x)}$ .

Any metric space is a fortiori a semimetric space. In particular, ${\displaystyle \mathbb {R} ^{k}}$ , the ${\displaystyle k}$ -dimensional Euclidean space, is the canonical metric space in distance geometry.

The triangle inequality is omitted in the definition, because we do not want to enforce more constraints on the distances ${\displaystyle d_{ij}}$  than the mere requirement that they be positive.

In practice, semimetric spaces naturally arises from inaccurate measurements. For example, given three points ${\displaystyle A,B,C}$  on a line, with ${\displaystyle d_{AB}=1,d_{BC}=1,d_{AC}=2}$ , an inaccurate measurement could give ${\displaystyle d_{AB}=0.99,d_{BC}=0.98,d_{AC}=2.00}$ , violating the triangle inequality.

#### Isometric embedding

Given two semimetric spaces, ${\displaystyle (R,d),(R',d')}$ , an isometric embedding from ${\displaystyle R}$  to ${\displaystyle R'}$  is a map ${\displaystyle f:R\to R'}$  that preserves the semimetric, that is, for all ${\displaystyle x,y\in R}$ , ${\displaystyle d(x,y)=d'(f(x),f(y))}$ .

For example, given the finite semimetric space ${\displaystyle (R,d)}$  defined above, an isometric embedding into is defined by points ${\textstyle A_{0},A_{1},\ldots ,A_{n}\in {\displaystyle \mathbb {R} ^{k}}}$ , such that ${\displaystyle d(A_{i},A_{j})=d_{ij}}$  for all ${\displaystyle 0\leq i .

#### Affine independence

Given the points ${\textstyle A_{0},A_{1},\ldots ,A_{n}\in {\displaystyle \mathbb {R} ^{k}}}$ , they are defined to be affinely independent, iff they cannot fit inside a single ${\displaystyle l}$ -dimensional affine subspace of ${\displaystyle \mathbb {R} ^{k}}$ , for any ${\displaystyle l , iff the ${\displaystyle n}$ -simplex they span, ${\displaystyle v_{n}}$ , has positive ${\displaystyle n}$ -volume, that is, ${\displaystyle Vol_{n}(v_{n})>0}$ .

In general, when ${\displaystyle k\geq n}$ , they are affinely independent, since a generic n-simplex is nondegenerate. For example, 3 points in the plane, in general, are not collinear, because the triangle they span does not degenerate into a line segment. Similarly, 4 points in space, in general, are not coplanar, because the tetrahedron they span does not degenerate into a flat triangle.

When ${\displaystyle n>k}$ , they must be affinely dependent. This can be seen by noting that any ${\displaystyle n}$ -simplex that can fit inside ${\displaystyle \mathbb {R} ^{k}}$  must be "flat".

## Cayley–Menger determinants

Cayley–Menger determinants, named after Arthur Cayley and Karl Menger, are determinants of matrices of distances between sets of points.

Let ${\textstyle A_{0},A_{1},\ldots ,A_{n}}$  be n + 1 points in a semimetric space, their Cayley–Menger determinant is defined by

${\displaystyle CM(A_{0},\cdots ,A_{n})={\begin{vmatrix}0&d_{01}^{2}&d_{02}^{2}&\cdots &d_{0n}^{2}&1\\d_{01}^{2}&0&d_{12}^{2}&\cdots &d_{1n}^{2}&1\\d_{02}^{2}&d_{12}^{2}&0&\cdots &d_{2n}^{2}&1\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\d_{0n}^{2}&d_{1n}^{2}&d_{2n}^{2}&\cdots &0&1\\1&1&1&\cdots &1&0\end{vmatrix}}}$

If ${\textstyle A_{0},A_{1},\ldots ,A_{n}\in {\displaystyle \mathbb {R} ^{k}}}$ , then they make up the vertices of an (possibly degenerate) n-simplex ${\displaystyle {\displaystyle v_{n}}}$  in ${\displaystyle \mathbb {R} ^{k}}$ . It can be shown that[6] the n-dimensional volume of the simplex ${\displaystyle {\displaystyle v_{n}}}$  satisfies

${\displaystyle Vol_{n}(v_{n})^{2}={\frac {(-1)^{n+1}}{(n!)^{2}2^{n}}}CM(A_{0},\cdots ,A_{n})}$ .

Note that, for the case of ${\displaystyle n=0}$ , we have ${\displaystyle Vol_{0}(v_{0})=1}$ , meaning the "0-dimensional volume" of a 0-simplex is 1, that is, there is 1 point in a 0-simplex.

${\textstyle A_{0},A_{1},\ldots ,A_{n}}$  are affinely independent iff ${\displaystyle Vol_{n}(v_{n})>0}$ , that is, ${\displaystyle (-1)^{n+1}CM(A_{0},\cdots ,A_{n})>0}$ . Thus Cayley–Menger determinants give a computational way to prove affine independence.

If ${\displaystyle k , then the points must be affinely dependent, thus ${\displaystyle CM(A_{0},\cdots ,A_{n})=0}$ . Cayley's 1841 paper studied the special case of ${\displaystyle k=3,n=4}$ , that is, any 5 points ${\displaystyle A_{0},\cdots ,A_{5}}$  in 3-dimensional space must have ${\displaystyle CM(A_{0},\cdots ,A_{4})=0}$ .

## History

The first result in distance geometry is Heron's formula, from 1st century AD, which gives the area of a triangle from the distances between its 3 vertices. Brahmagupta's formula, from 7th century AD, generalizes it to cyclic quadrilaterals. Tartaglia, from 16th century AD, generalized it to give the volume of tetrahedron from the distances between its 4 vertices.

The modern theory of distance geometry began with Authur Cayley and Karl Menger.[7] Cayley published the Cayley determinant in 1841[8], which is a special case of the general Cayley–Menger determinant. Menger proved in 1928 a characterization theorem of all semimetric spaces that are isometrically embeddable in the n-dimensional Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ .[9][10] In 1931, Menger used distance relations to give an axiomatic treatment of Euclidean geometry.[11]

Leonard Blumenthal's book[12] gives a general overview for distance geometry at the graduate level, a large part of which is treated in English for the first time when it was published.

## Menger characterization theorem

Menger proved the following characterization theorem of semimetric spaces:[2]

A semimetric space ${\displaystyle (R,d)}$  is isometrically embeddable in the ${\displaystyle n}$ -dimensional Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ , but not in ${\displaystyle \mathbb {R} ^{m}}$  for any ${\displaystyle 0\leq m , if and only if:

1. ${\displaystyle R}$  contains an ${\displaystyle (n+1)}$ -point subset ${\displaystyle S}$  that is isometric with an affinely independent ${\displaystyle (n+1)}$ -point subset of ${\displaystyle \mathbb {R} ^{n}}$ ;
2. any ${\displaystyle (n+3)}$ -point subset ${\displaystyle S'}$ , obtained by adding any two additional points of ${\displaystyle R}$  to ${\displaystyle S}$ , is congruent to an ${\displaystyle (n+3)}$ -point subset of ${\displaystyle \mathbb {R} ^{n}}$ .

A proof of this theorem in a slightly weakened form (for metric spaces instead of semimetric spaces) is in [13].

## Characterization via Cayley–Menger determinants

The following results are proved in Blumethal's book[12].

### Embedding ${\displaystyle n+1}$  points in ${\displaystyle \mathbb {R} ^{n}}$

Given a semimetric space ${\displaystyle (S,d)}$  , with ${\displaystyle S=\{P_{0},\cdots ,P_{n}\}}$ , and ${\displaystyle d(P_{i},P_{j})=d_{ij}\geq 0}$ , ${\displaystyle 0\leq i , an isometric embedding of ${\displaystyle (S,d)}$  into ${\displaystyle \mathbb {R} ^{n}}$  is defined by ${\textstyle A_{0},A_{1},\ldots ,A_{n}\in {\displaystyle \mathbb {R} ^{n}}}$ , such that ${\displaystyle d(A_{i},A_{j})=d_{ij}}$  for all ${\displaystyle 0\leq i .

Again, one asks whether such an isometric embedding exists for ${\displaystyle (S,d)}$ .

A necessary condition is easy to see: for all ${\displaystyle k=1,\cdots ,n}$ , let ${\displaystyle v_{k}}$  be the k-simplex formed by ${\textstyle A_{0},A_{1},\ldots ,A_{k}}$ , then

${\displaystyle (-1)^{k+1}CM(P_{0},\cdots ,P_{k})=(-1)^{k+1}CM(A_{0},\cdots ,A_{k})=2^{k}(k!)^{k}Vol_{k}(v_{k})^{2}\geq 0}$

The converse also holds. That is, if for all ${\displaystyle k=1,\cdots ,n}$ ,

${\displaystyle (-1)^{k+1}CM(P_{0},\cdots ,P_{k})\geq 0}$ ,

then such an embedding exists.

Further, such embedding is unique up to isometry in ${\displaystyle \mathbb {R} ^{n}}$ . That is, given any two isometric embeddings defined by ${\textstyle A_{0},A_{1},\ldots ,A_{n}}$ , and ${\textstyle A'_{0},A'_{1},\ldots ,A'_{n}}$ , there exists a (not necessarily unique) isometry ${\displaystyle T:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}$ , such that ${\displaystyle T(A_{k})=A'_{k}}$  for all ${\displaystyle k=0,\cdots ,n}$ . Such ${\displaystyle T}$  is unique if and only if ${\displaystyle CM(P_{0},\cdots ,P_{n})\neq 0}$ , that is, ${\textstyle A_{0},A_{1},\ldots ,A_{n}}$  are affinely independent.

### Embedding ${\displaystyle n+2}$  and ${\displaystyle n+3}$  points

If ${\displaystyle n+2}$  points ${\displaystyle P_{0},\cdots ,P_{n+1}}$  can be embedded in ${\displaystyle \mathbb {R} ^{n}}$  as ${\displaystyle A_{0},\cdots ,A_{n+1}}$ , then other than the conditions above, an additional necessary condition is that the ${\displaystyle (n+1)}$ -simplex formed by ${\textstyle A_{0},A_{1},\ldots ,A_{n+1}}$ , must have no ${\displaystyle (n+1)}$ -dimensional volume. That is, ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1})=0}$ .

The converse also holds. That is, if for all ${\displaystyle k=1,\cdots ,n}$ ,

${\displaystyle (-1)^{k+1}CM(P_{0},\cdots ,P_{k})\geq 0}$ ,

and

${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1})=0}$ ,

then such an embedding exists.

For embedding ${\displaystyle n+3}$  points in ${\displaystyle \mathbb {R} ^{n}}$ , the necessary and sufficient conditions are similar:

1. For all ${\displaystyle k=1,\cdots ,n}$ , ${\displaystyle (-1)^{k+1}CM(P_{0},\cdots ,P_{k})\geq 0}$ ;
2. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1})=0}$ ;
3. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+2})=0}$ ;
4. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1},P_{n+2})=0}$ .

### Embedding arbitrarily many points

The ${\displaystyle n+3}$  case turns out to be sufficient in general.

In general, given a semimetric space ${\displaystyle (R,d)}$ , it can be isometrically embedded in ${\displaystyle \mathbb {R} ^{n}}$  if and only if there exists ${\displaystyle P_{0},\cdots ,P_{n}\in R}$ , such that, for all ${\displaystyle k=1,\cdots ,n}$ , ${\displaystyle (-1)^{k+1}CM(P_{0},\cdots ,P_{k})\geq 0}$ , and for any ${\displaystyle P_{n+1},P_{n+2}\in R}$ ,

1. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1})=0}$ ;
2. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+2})=0}$ ;
3. ${\displaystyle CM(P_{0},\cdots ,P_{n},P_{n+1},P_{n+2})=0}$ .

And such embedding is unique up to isometry in ${\displaystyle \mathbb {R} ^{n}}$ .

Further, if ${\displaystyle CM(P_{0},\cdots ,P_{n})\neq 0}$ , then it cannot be isometrically embedded in any ${\displaystyle \mathbb {R} ^{m},m . And such embedding is unique up to unique isometry in ${\displaystyle \mathbb {R} ^{n}}$ .

Thus, Cayley–Menger determinants give a concrete way to calculate whether a semimetric space can be embedded in ${\displaystyle \mathbb {R} ^{n}}$ , for some finite ${\displaystyle n}$ , and if so, what is the minimal ${\displaystyle n}$ .

## Applications

There are many applications of distance geometry.[3]

In telecommunication networks such as GPS, the positions of some sensors are known (which are called anchors) and some of the distances between sensors are also known: the problem is to identify the positions for all sensors.[5] Hyperbolic navigation is one pre-GPS technology that uses distance geometry for locating ships based on the time it takes for signals to reach anchors.

There are many applications in chemistry.[4][12] Techniques such as NMR can measure distances between pairs of atoms of a given molecule, and the problem is to infer the 3-dimensional shape of the molecule from those distances.

Some software packages for applications are:

## References

1. ^ Yemini, Y. (1978). "The positioning problem — a draft of an intermediate summary". Conference on Distributed Sensor Networks, Pittsburgh.
2. ^ a b Liberti, Leo; Lavor, Carlile; MacUlan, Nelson; Mucherino, Antonio (2014). "Euclidean Distance Geometry and Applications". SIAM Review. 56: 3–69. arXiv:1205.0349. doi:10.1137/120875909.
3. ^ a b Mucherino, A.; Lavor, C.; Liberti, L.; Maculan, N. (2013). Distance Geometry: Theory, Methods and Applications.
4. ^ a b c Crippen, G.M.; Havel, T.F. (1988). Distance Geometry and Molecular Conformation. John Wiley & Sons.
5. ^ a b Biswas, P.; Lian, T.; Wang, T.; Ye, Y. (2006). "Semidefinite programming based algorithms for sensor network localization". ACM Transactions on Sensor Networks. 2 (2): 188–220. doi:10.1145/1149283.1149286.
6. ^ "Simplex Volumes and the Cayley-Menger Determinant". www.mathpages.com. Archived from the original on 16 May 2019. Retrieved 2019-06-08.
7. ^ Liberti, Leo; Lavor, Carlile (2016). "Six mathematical gems from the history of distance geometry". International Transactions in Operational Research. 23 (5): 897–920. arXiv:1502.02816. doi:10.1111/itor.12170. ISSN 1475-3995.
8. ^ Cayley, Arthur (1841). "On a theorem in the geometry of position". Cambridge Mathematical Journal. 2: 267–271.
9. ^ Menger, Karl (1928-12-01). "Untersuchungen über allgemeine Metrik". Mathematische Annalen (in German). 100 (1): 75–163. doi:10.1007/BF01448840. ISSN 1432-1807.
10. ^ Blumenthal, L. M.; Gillam, B. E. (1943). "Distribution of Points in n-Space". The American Mathematical Monthly. 50 (3): 181. doi:10.2307/2302400. JSTOR 2302400.
11. ^ Menger, Karl (1931). "New Foundation of Euclidean Geometry". American Journal of Mathematics. 53 (4): 721–745. doi:10.2307/2371222. ISSN 0002-9327. JSTOR 2371222.
12. ^ a b c Blumenthal, L.M. (1970). Theory and applications of distance geometry (2nd ed.). Bronx, New York: Chelsea Publishing Company. pp. 90–161. ISBN 978-0-8284-0242-2. LCCN 79113117.
13. ^ Bowers, John C.; Bowers, Philip L. (2017-12-13). "A Menger Redux: Embedding Metric Spaces Isometrically in Euclidean Space". The American Mathematical Monthly. 124 (7): 621. doi:10.4169/amer.math.monthly.124.7.621. S2CID 50040864.


text:
## anonymous one year ago Someone please help me with this problem, I need to show my work and I have no idea how or where to start : Find the cube roots of 27(cos 330° + i sin 330°).

1. IrishBoy123

you know $$e^{i \theta} = cos \theta + i \ sin \theta$$??

2. anonymous

I've seen it before in my lessons but I have never actually applied it to an problems

3. IrishBoy123

it's is very useful to know you can write $$27(cos 330° + i sin 330°)$$ as $$\large 27 \ e^{i \ \frac{11}{6}\pi}$$ so you want $$\large \sqrt[3] { 27 \ e^{i \ \frac{11}{6}\pi}}$$

4. anonymous

That looks very confusing to solve Im uncertain as to how im supposed to take the cube root of that number

5. IrishBoy123

there's something related called deMoivre, which might be easier for you right now $$(cos \theta + i \ sin \theta)^n = cos \ n \ \theta + i \ sin \ n \ \theta$$

6. anonymous

Yes, I've heard of DeMoivre's Theorem

7. IrishBoy123

for you $$n = \frac{1}{3}$$ and don't forget the $$\sqrt[3] 27$$ !

8. anonymous

So $(\cos \theta+i \sin \theta)^{\frac{ 1 }{ 3 }}=\cos (\frac{ 1 }{ 3 })\theta+i \sin (\frac{ 1 }{ 3 })\theta ?$

9. IrishBoy123

not that simple! you will get roots following this pattern $$27^{1/3} \ cis(\frac{2π \ m+\theta }{3})$$ for m = 0, 1, 2

10. IrishBoy123

ie 3 roots for a cubic

11. anonymous

What is $cis(\frac{ 2pim+\theta }{ 3 }) ???$ Im confused. I know there will be 3 answers to this because its asking for cube roots.

12. IrishBoy123

sorry to confuse you, cis is just a shorthand for $$cos + i sin$$ so stuff that argument into both of them

13. anonymous

What does the p stand for in that above formula?

14. anonymous

I believe he meant pi times m, it just put pim instead of writing the letter for pi

15. IrishBoy123

so you will have $$3(cos \frac{\theta}{3} + i sin \frac{\theta }{3})$$ $$3(cos \frac{\theta + 2 \pi}{3} + i sin \frac{\theta + 2 \pi}{3})$$ $$3(cos \frac{\theta + 4 \pi}{3} + i sin \frac{\theta + 4 \pi}{3})$$

16. anonymous

$\pi m$ not pim

17. anonymous

is what he meant

18. IrishBoy123

from deMoivre, i think you have already seen how to get the first root ie when you wrote : $$(cosθ+isinθ)^{1/3}=cos(1/3)θ+isin(1/3)θ$$ the other roots come up only because the sinusoid repeats itself with period $$2 \pi$$ and you need to look at all solutions for a cube root, the roots will repeat themselves after you have the first 3. for, say, sixth root, you will get six distinct roots and then they will repeat.... and don;t forget the $$\sqrt[3]{27}$$ ....


text:
# Efficiently Computing the Determinant of a Matrix, Part 2: Determinant by Elementary Row Operations

By Nathan Reynoso on

Note: This post is part 2 of a 2-part series: part 1, part 2.

Computing the determinant by hand is often annoying, and using the cofactor method takes more time the larger the matrix. The most efficient way to compute the determinant of a matrix is through using elementary row operations. Through this process, we reduce the matrix to echelon form and take the operations we did on the rows to compute the determinant.

Our first step is to take any given matrix to echelon form, but we have to keep track of the row operations we apply to be able to use them later when computing the determinant. Below, I have a randomly generated a $3\times 3$ matrix:

\begin{align*} \begin{bmatrix} 0 & 2 & 4\\ 7 & 24 & 1\\ 2 & 6 & 8 \end{bmatrix} \end{align*}

The frst thing we would do is switch our top and bottom rows ($R_1$ and $R_3$). Then, we would divide our top row by 2 to get our pivot row for column $1$. We’ll need to keep a list of all numbers we divide any row by and the number of times we switch rows.

\begin{align*} \begin{bmatrix} 1&3&4\\ 7&24&1\\ 0&2&4 \end{bmatrix} , \text{row divisors:(2), row swaps:(1)} \end{align*}

Next we subtract $7\times R_{1}$ from $R_{2}$. After that we subtract $R_{2}$ from $R_{1}$.

\begin{align*} \begin{bmatrix} 1&3&4\\ 0&3&20\\ 0&2&4 \end{bmatrix}, \text{row divisors:(2), row swaps:(1)} \end{align*}

\begin{align*} \begin{bmatrix} 1&0&-16\\ 0&3&-27\\ 0&2&4 \end{bmatrix}, \text{row divisors:(2), row swaps:(1)} \end{align*}

Now we divide $R_{2}$ by 3, and subtract $2\times R_{2}$ from $R_{3}$. We can also divide $R_{3}$ by $22$ as well.

\begin{align*} \begin{bmatrix} 1&0&-16\\ 0&1&-9\\ 0&0&1 \end{bmatrix},\text{row divisors:(2,3,22), row swaps:(1)} \end{align*}

At this point we can see the matrix does reduce to the identity, so we can stop here. Now with the row divisors and swaps we collected we can use them to calculate the determinant. To get it, we take the product of the divisors and multiply it by $-1$ to the power of the number of row swaps:

\begin{align*} (2\times 3\times 22)\times -1^{1} = -132 \end{align*}

When computing the determinant using either the cofactor method or by using a calculator, we can see this is actually the determinant.

## Implementation

Below is the actual code for computing the determinant via row operations that I implemented into my matrix class. The first thing it’s doing is making a copy of the matrix, so that we don’t mutate the original matrix. Then we make a variable called factors that will be what the method returns, and that’s what we multiply our dividing numbers by.

After, we ensure the matrix is square, otherwise you can’t compute the determinant. We then loop throw each row of the matrix, checking which row is the pivot row, and making sure the pivot row is where its supposed to be (in the position of the next column that needs a pivot). Once we have the pivot row in the correct position, we multiply the factors variable by the pivot number, and then reduce the column that number is in to $1$’s and $0$’s. Then we move on to the next row. After going through the whole matrix, we return the factors variable.

def calc_determinant(self):
matrix = self.copy()
factors = 1
row_index = 0

if matrix.num_cols != matrix.num_rows:
return ("Non-square matrices have no determinant.")

for col in range(matrix.num_cols):
pivot = matrix.get_pivot_row(col)

if pivot != None:
if pivot != row_index:
matrix = matrix.swap_rows(row_index, pivot)
factors *= -1

factors *= matrix.elements[pivot][col]
matrix = matrix.normalize_row(row_index)
matrix = matrix.clear_above(row_index)
matrix = matrix.clear_below(row_index)
row_index += 1

else:
mult_constant *= 0
continue

return factors

## Time Complexity

When using this method to compute the determinant, the computation is relatively fast compared to other methods. Lets assume we have a $n \times n$ matrix that can be taken to echelon form, then we can easily find the maximum amount of computations needed to find the determinant using this method. Assuming that all the entries are not $1$, then for each row, you would divide by a number at some point, so that’s $n$ computations so far. For each row, you will also have to subtract it from all the other rows to make it the pivot row, which is $n-1$ more computations for $n$ rows, which is $n(n-1)$. This gives us $n+n(n-1) = n^2$ total computations.

This is much more efficient than computing the determinant via the cofactor method, which can take $n!$ operations.

Acknowledgements: Thanks to Charlie Weinberger for reviewing this post.


text:
# The Mathematics of Juggling and more from George Hart

[Dr. Chase guest blogging again]

You’re probably familiar with Vi Hart’s math videos. Less well-known are her father’s math videos. Although I was aware of his mathematical sculpture, I was not aware until today that since August 2012, he has been producing a mathematical video series called mathematical impressions for the Simons Foundation. The 10th in the series is The Mathematics of Juggling. Check it out!

# Progress Toward Twin-Prime Conjecture

This nice article came through on wired today:

# Unknown Mathematician Proves Surprising Property of Prime Numbers

By Erica Klarreich, Simons Science News

Image: bwright923/Flickr

On April 17, a paper arrived in the inbox of Annals of Mathematics, one of the discipline’s preeminent journals. Written by a mathematician virtually unknown to the experts in his field — a 50-something lecturer at the University of New Hampshire named Yitang Zhang — the paper claimed to have taken a huge step forward in understanding one of mathematics’ oldest problems, the twin primes conjecture.

Editors of prominent mathematics journals are used to fielding grandiose claims from obscure authors, but this paper was different. Written with crystalline clarity and a total command of the topic’s current state of the art, it was evidently a serious piece of work, and the Annals editors decided to put it on the fast track.

Just three weeks later — a blink of an eye compared to the usual pace of mathematics journals — Zhang received the referee report on his paper.

“The main results are of the first rank,” one of the referees wrote. The author had proved “a landmark theorem in the distribution of prime numbers.”

(more)

This is very exciting news, and the whole story has a fantastic David & Goliath feel–“little known mathematician delivers a crushing blow to a centuries old problem” (not a fatal blow, but a crushing one). It’s such a feel-good story, almost like Andrew Wiles and Fermat’s Last Theorem. Here’s my favorite part of the article:

…during a half-hour lull in his friend’s backyard before leaving for a concert, the solution suddenly came to him. “I immediately realized that it would work,” he said.

Just chillin’ in his friend’s backyard…and it came to him! Anyone who has worked on math problems or puzzles has had this experience, right? It seems like an experience common to all people. This has definitely happened to me lots of times–an insight hits me out of nowhere and unlocks a problem I’ve been working on for weeks. It’s one of the reasons we do mathematics!

# What is a Point of Inflection?

Simple question right?

This website, along with the Calc book we’re teaching from, define it this way:

A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.

No debate about there being an inflection point at x=0 on this graph.

There’s no debate about functions like $f(x)=x^3-x$, which has an unambiguous inflection point at $x=0$.

In fact, I think we’re all in agreement that:

1. There has to be a change in concavity. That is, we require that for $x we have $f''(x)<0$ and for $x>c$ we have $f''(x)>0$, or vice versa.*
2. The original function $f$ has to be continuous at $x=c$. That is, $f(x)=\frac{1}{x^2}$ does not have a point of inflection at $x=0$ even though there’s a concavity change because $f$ isn’t even defined here. If we then piecewise-define $f$ so that it carries the same values except at $x=0$ for which we define $f(0)=5$, we still don’t consider this a point of inflection because of the lack of continuity.

The point of inflection x=0 is at a location without a first derivative. A “tangent line” still exists, however.

But the part of the definition that requires $f$ to have a tangent line is problematic, in my opinion. I know why they say it this way, of course. They want to capture functions that have a concavity change across a vertical tangent line, such as $f(x)=\sqrt[3]{x}$. Here we have a concavity change (concave up to concave down) across $x=0$ and there is a tangent line ($x=0$) but $f'(0)$ is undefined.

Is x=0 a point of inflection? Some definitions say no, because no tangent line exists.

So It’s clear that this definition is built to include vertical tangents. It’s also obvious that the definition is built in such a way as to exclude cusps and corners. Why? What’s wrong with a cusp or corner being a point of inflection? I would claim that the piecewise-defined function $f(x)$ shown above has a point of inflection at $x=0$ even though no tangent line exists here. [Edit: Originally I had used $f(x)=x^{2/3}$ as my example, but as my dad so astutely pointed out, that does have a tangent line at zero. Doh! Thanks dad!]

I prefer the definition:

A point where the graph of a function is continuous and where the concavity changes is a point of inflection.

That is, I would only require the two conditions listed at the beginning of this post. What do you think?

Once you’re done thinking about that, consider this strange example that has no point of inflection even though there’s a concavity change. As my colleague Matt suggests, could we consider this a region of inflection? Now we’re just being silly, right?

A region/interval of inflection?

———————————————

Footnotes:

* When we say that a function is concave up or down on a certain interval, we actually mean $f''(x)>0$ or $f''(x)<0$ for the whole interval except at finitely many locations. If there are point discontinuities, we still consider the interval to have the same concavity.

** This source, interestingly, seems to require differentiability at the point. I think most of us would agree this is too strong a requirement, right?

# Friday fun from around the web

Here are two fun mathy things that came through my feed today. Many of you have probably already seen today’s math-themed xkcd:

And I also saw this today [on thereifixedit], which delighted the mathematician in me:

Happy Friday everyone!

I’ve been loving the videos that SpikedMathGames has been posting on youtube. Check out their channel here. In particular, I’ve enjoyed Paradox Tuesday. Here’s one from a few weeks ago which really interested me (if you go to the youtube page, you’ll see I’ve been active in the comments!):

They also cover some famous paradoxes like the Unexpected Hanging Paradox or the  Barber Paradox. If you’ve never heard of these, go watch these now.

I’m especially interested in paradoxes that deal with infinity, countability, and probability. Here’s another great paradox that deals with just those issues that my friend Matthew Wright shared with me a few months ago (thanks Matthew!). It’s called the Grim Reaper paradox (can’t link to the Wikipedia article–it doesn’t yet exist), proposed in 1964 by José Amado Benardete in his book Infinity: an essay in metaphysics, and I first read about it on Alexander Pruss’s blog here, and I quote:

Say that a Grim Reaper is a being that has the following properties: It wakes up at a time between 8 and 9 am, both exclusive, and if you’re alive, it instantaneously kills you, and if you’re not alive, it doesn’t do anything. Suppose there are countably infinitely many Grim Reapers, and before they go to bed for the night, each sets his alarm for a time (not necessarily the same time as the other Reapers) strictly between 8 and 9 am. Suppose, also, that no other kind of death is available for you, and that you’re not going to be resurrected that day.

Then, you’re going to be dead at 9 am, since as long as at least one Grim Reaper wakes up during that time period, you’re guaranteed to be dead. Now whether there is a paradox here depends on how the Grim Reapers individually set their alarm clocks. Suppose now that they set them in such a way that the following proposition p is true:

(p) for every time t later than 8 am, at least one of the Grim Reapers woke up strictly between 8 am and t.

Here’s a useful Theorem: If the Grim Reapers choose their alarm clock times independently and uniformly over the 8-9 am interval, then P(p)=1.

Now, if p is true, then no Grim Reaper kills you. For suppose that a Grim Reaper who wakes up at some time t1, later than 8 am, kills you. If p is true, there is a Grim Reaper who woke up strictly between 8 am and t1, say at t0. But if so, then you’re going to be dead right after t0, and hence the Grim Reaper who woke up at t1 is not going to do anything, since you’re dead then. Hence, if p is true, no Grim Reaper kills you. On the other hand, I’ve shown that it is certain that a Grim Reaper kills you. Hence, if p is true, then no Grim Reaper kills you and a Grim Reaper kills you, which is absurd.

Go visit his blog post for a discussion of why this seems unresolvable, and how it may actually put forward a case for time being discrete rather than continuous. Crazy thought.

I sometimes ask my students this somewhat related question–perhaps you’ve heard it too:

How many positive integers have a 3 in them? (That is, in their decimal representation. 6850104302 has a 3 but 942009947 does not.)

If you haven’t ever considered this question, take the time to do it now.

Though I actually once worked out the result using limits (like Alexander Bogomolny does marvelously here), it’s easy enough to work out the result in our heads:

First ask yourself how many digits a randomly selected integer has. The number of digits is almost certainly greater than 2, right? There are only 90 two-digit positive integers, a finite number, and there are an infinite number of integers with more than two digits. It follows that if you were to pick one at random from among all positive integers*, it would be almost certain to contain more than two digits.

The same argument could be applied to a larger number of digits. By the same logic as above, we can convince ourselves that ‘most randomly selected integers have more than a trillion digits’. It’s a bit of an incredible statement, really. We rarely ever work with the ‘most-common’ kind of numbers (the big ones!).**

What is the probability that a number with a trillion digits has a 3 in it? Well, it’s almost certain. The probability approaches 100%. If we consider ALL numbers, the probability IS 100% (or is it?). This is a real dilemma. How can we say that 100% of numbers have a 3 in them when this is clearly not true?

We’ve been pretty sloppy here, but regardless, this kind of fast-and-loose infinite probability question is unsettling.

Do you want to try taking a crack at these? Feel free to comment below.

Oh, and Happy Birthday Euler!

_________________________________

# Flat Donuts

[Guest blogger Dr. Gene Chase]

You know about flat donuts if you played a computer game in which when you go off the screen at the left, you return coming in at the right, and similarly when you go off the screen at the top, you return coming in at the bottom.

Mathematicians call the rectangle of your computer screen a flat torus. The word “torus” reminds us that we are only thinking of the surface of the donut.

What does that computer screen have to do with a torus? Stretch the screen around a torus as this picture begins to show.

Since the left and right sides of the screen are “the same” and the top and bottom sides of the screen are “the same,” the screen seamlessly takes the shape of the donut. Mathematicians say that the flat donut and the donut have the same topology, because we bent one into the other without cutting or pasting. (I remind you that opposite edges of the flat donut were already pasted by regarding them as “the same” before we began to bend it.)

Although a flat torus and a torus have the same topology, they do not have the same geometry. Geometry is about measuring space. In particular, on a flat torus, the shortest distance between two points is always a straight line. But on a torus, the shortest distance between two points staying in the torus is never a straight line.

Is it possible to paste the opposite edges of the flat torus together in such a way that the resulting thing in 3D has the same geometry? That is, such that straight lines are still straight?

It is easy to take one step, to create a cylinder, by pasting together just one pair of opposite edges. Notice that no stretching is involved at all.

How about pasting both pairs of edges? My intuition says that such a thing is impossible.

My intuition is wrong. In 1954, John Nash — yes, that John Nash of A Beautiful Mind — proved that it is possible, but without saying how. (He gave a so-called “existence proof.”)

But only 11 months ago did we have a picture of what the resulting flat torus would look like. Here’s a news article with a computer-generated picture to illustrate it.


text: Ucci, John Three remarks on symmetric products and symmetric maps Pacific Journal of Mathematics 45 (1973) 369–377 Uchiyama, Akihito Extension of the Hardy-Littlewood-Fefferman-Stein inequality Pacific Journal of Mathematics 120 (1985) 229–255 On McConnell's inequality for functionals of subharmonic functions Pacific Journal of Mathematics 128 (1987) 367–377 On the radial maximal function of distributions Pacific Journal of Mathematics 121 (1986) 467–483 The construction of certain BMO functions and the corona problem Pacific Journal of Mathematics 99 (1982) 183–204 The factorization of $H^{p}$ on the space of homogeneous type Pacific Journal of Mathematics 92 (1981) 453–468 The Fefferman-Stein decomposition of smooth functions and its application to $H^{p}(\mathbf{R}^{n})$ Pacific Journal of Mathematics 115 (1984) 217–255 Uchiyama, Mitsuru Commutativity of selfadjoint operators Pacific Journal of Mathematics 161 (1993) 385–392 Uda, Toshio Complex bases of certain semiproper holomorphic maps Pacific Journal of Mathematics 78 (1978) 549–552 Ueda, Yoshimichi Amalgamated free product over Cartan subalgebra Pacific Journal of Mathematics 191 (1999) 359–392 Uehara, Tsuyoshi On class numbers of cyclic quartic fields Pacific Journal of Mathematics 122 (1986) 251–255 Ueyama, Kenta Cluster tilting modules and noncommutative projective schemes Pacific Journal of Mathematics 289 (2017) 449–468 Ugarte, Luis Strong Kähler with torsion structures from almost contact manifolds Pacific Journal of Mathematics 249 (2011) 49–75 Uhl, John Martingales of vector valued set functions Pacific Journal of Mathematics 30 (1969) 533–548 Norm attaining operators on $L^1[0,1]$ and the Radon-Nikodým property Pacific Journal of Mathematics 63 (1976) 293–300 Radon-Nikodým theorems for the Bochner and Pettis integrals Pacific Journal of Mathematics 38 (1971) 531–536 Vector-valued functions as families of scalar-valued functions Pacific Journal of Mathematics 95 (1981) 75–83 Uhlig, Frank Definite and semidefinite matrices in a real symmetric matrix pencil Pacific Journal of Mathematics 49 (1973) 561–568 On the maximal number of linearly independent real vectors annihilated simultaneously by two real quadratic forms Pacific Journal of Mathematics 49 (1973) 543–560 The number of vectors jointly annihilated by two real quadratic forms determines the inertia of matrices in the associated pencil Pacific Journal of Mathematics 49 (1973) 537–542 Ülger, A. Arens regularity sometimes implies the RNP Pacific Journal of Mathematics 143 (1990) 377–399 Ullian, Joseph Approximation of functions on the integers Pacific Journal of Mathematics 13 (1963) 693–701 Ullman, Daniel A generalization of a theorem of Atkinson to noninvariant measures Pacific Journal of Mathematics 130 (1987) 187–193 Ullman, Joseph On Tchebycheff polynomials Pacific Journal of Mathematics 9 (1959) 913–923 Ullom, Stephen Minimal ramification in nilpotent extensions Pacific Journal of Mathematics 253 (2011) 125–143 Ulmer, Douglas On the arithmetic of a family of twisted constant elliptic curves Pacific Journal of Mathematics 305 (2020) 597–640 Ulmer, Milton $C$-embedded $\Sigma$-spaces Pacific Journal of Mathematics 46 (1973) 591–602 Umehara, Masaaki Addendum: singularities of flat fronts in hyperbolic space Pacific Journal of Mathematics 294 (2018) 505–509 C^1-umbilics with arbitrarily high indices Pacific Journal of Mathematics 288 (2017) 1–26 Flat fronts in hyperbolic 3-space Pacific Journal of Mathematics 216 (2004) 149–175 Singularities of flat fronts in hyperbolic space Pacific Journal of Mathematics 221 (2005) 303–351 Uneme, Yuya Angular distribution of diameters for spheres and rays for planes Pacific Journal of Mathematics 277 (2015) 169–185 Ungar, Abraham A. Gyrogroups and the decomposition of groups into twisted subgroups and subgroups Pacific Journal of Mathematics 197 (2001) 1–11 Ungar, Gerald Conditions for a mapping to have the slicing structure property Pacific Journal of Mathematics 30 (1969) 549–553 Unsurangsie, Sumalee A semilinear wave equation with nonmonotone nonlinearity Pacific Journal of Mathematics 132 (1988) 215–225 Uraltsev, Gennady On the distributional Hessian of the distance function Pacific Journal of Mathematics 270 (2014) 151–166 Urbano, Francisco Closedconformal vector fields and Lagrangian submanifolds in complex space forms Pacific Journal of Mathematics 199 (2001) 269–302 Hamiltonian-minimal lagrangian submanifolds in complex space forms Pacific Journal of Mathematics 227 (2006) 43–63 Urbano-Blanco, J. Correction to:\\Modular Diophantine inequalities\\and numerical semigroups Pacific Journal of Mathematics 220 (2005) 199–199 Modular diophantine inequalities and numerical semigroups Pacific Journal of Mathematics 218 (2005) 379–398 Urbański, Mariusz Conformal repellors with dimension one are Jordan curves Pacific Journal of Mathematics 166 (1994) 85–97 Hausdorff dimension of limit sets for parabolic IFS with overlaps Pacific Journal of Mathematics 201 (2001) 441–478 Metrical diophantine analysis for tame parabolic iterated function systems Pacific Journal of Mathematics 216 (2004) 361–392 Uribe Jongbloed, Bernardo On the classification of pointed fusion categories up to weak Morita equivalence Pacific Journal of Mathematics 290 (2017) 437–466 Usachev, Alexandr Invariant Banach limits and applications to noncommutative geometry Pacific Journal of Mathematics 306 (2020) 357–373 Usami, Hiroyuki A Liouville type theorem for semilinear elliptic systems Pacific Journal of Mathematics 204 (2002) 247–255 Pairs of positive solutions of quasilinear elliptic equations in exterior domains Pacific Journal of Mathematics 120 (1985) 385–399 Usefi, Hamid Isomorphism invariants of restricted enveloping algebras Pacific Journal of Mathematics 246 (2010) 487–494 Ushijima, Akira Generic fundamental polygons for Fuchsian groups Pacific Journal of Mathematics 251 (2011) 453–468 Ustina, Fred The Hausdorff means of double Fourier series and the principle of localization Pacific Journal of Mathematics 37 (1971) 235–251 Uttersrud, U. Intersections of $M$-ideals and $G$-spaces Pacific Journal of Mathematics 104 (1983) 175–177 Utumi, Yuzo Intrinsic extensions of rings Pacific Journal of Mathematics 14 (1964) 505–512 Utz, Winfiled A note on unrestricted regular transformations Pacific Journal of Mathematics 1 (1951) 447–453 Minimizing integrals in certain classes of monotone functions Pacific Journal of Mathematics 7 (1957) 833–847 The non-existence of expansive homeomorphisms on a closed $2$-cell Pacific Journal of Mathematics 10 (1960) 1319–1321 Uyanik, Caglar Simultaneous construction of hyperbolic isometries Pacific Journal of Mathematics 294 (2018) 71–88


text:
# Weak convergence of an iterative algorithm for accretive operators

Volume 10, Issue 8, pp 4099--4108 Publication Date: August 07, 2017       Article History
• 532 Views

### Authors

Hengjun Zhao - School of Science, Henan University of Engineering, Zhengzhou 451191, China.
Sun Young Cho - Center for General Education, China Medical University, Taichung, Taiwan.

### Abstract

In this paper, an iterative algorithm investigated for $m$-accretive and inverse-strongly accretive operators. Also, a weak convergence theorem for the sum of two accretive operators is established in a real uniformly convex and $q$-uniformly smooth Banach space.

### Keywords

• Accretive operator
• zero point
• projection
• splitting method
• weak convergence.

•  47H06
•  90C33

### References

• [1] D. E. Alspach, A fixed point free nonexpansive map , Proc. Amer. Math. Soc., 82 (1981), 423–424.

• [2] I. K. Argyros, S. George , On the convergence of inexact Gauss-Newton method for solving singular equations, J. Nonlinear Funct. Anal., 2016 (2016), 22 pages.

• [3] B. A. Bin Dehaish, A. Latif, H. O. Bakodah, X.-L. Qin , A regularization projection algorithm for various problems with nonlinear mappings in Hilbert spaces, J. Inequal. Appl., 2015 (2015), 14 pages.

• [4] B. A. Bin Dehaish, X.-L. Qin, A. Latif, H. O. Bakodah , Weak and strong convergence of algorithms for the sum of two accretive operators with applications, J. Nonlinear Convex Anal., 16 (2015), 1321–1336.

• [5] E. Blum, W. Oettli, From optimization and variational inequalities to equilibrium problems, Math. Student, 63 (1994), 123–145.

• [6] F. E. Browder, Nonexpansive nonlinear operators in a Banach space, Proc. Nat. Acad. Sci. U.S.A., 54 (1965), 1041–1044.

• [7] R. E. Bruck, A simple proof of the mean ergodic theorem for nonlinear contractions in Banach spaces, Israel J. Math., 32 (1979), 107–116.

• [8] C. Byrne, A unified treatment of some iterative algorithms in signal processing and image reconstruction, Inverse Problems, 20 (2004), 103–120.

• [9] S. Y. Cho, B. A. Bin Dehaish, X.-L. Qin , Weak convergence of a splitting algorithm in Hilbert spaces, J. Appl. Anal. Comput., 7 (2017), 427–438.

• [10] S. Y. Cho, S. M. Kang, Approximation of common solutions of variational inequalities via strict pseudocontractions, Acta Math. Sci. Ser. B Engl. Ed., 32 (2012), 1607–1618.

• [11] S. Y. Cho, W.-L. Li, S. M. Kang, Convergence analysis of an iterative algorithm for monotone operators, J. Inequal. Appl., 2013 (2013), 14 pages.

• [12] J. García Falset, W. Kaczor, T. Kuczumow, S. Reich , Weak convergence theorems for asymptotically nonexpansive mappings and semigroups, Nonlinear Anal., 43 (2001), 377–401.

• [13] S.-P. Han, G. Lou , A parallel algorithm for a class of convex programs, SIAM J. Control Optim., 26 (1988), 345–355.

• [14] T. Kato, Nonlinear semigroups and evolution equations , J. Math. Soc. Japan, 19 (1967), 508–520.

• [15] S.-T. Lv , Convergence analysis of a Halpern-type iterative algorithm for zero points of accretive operators, Commun. Optim. Theory, 2016 (2016), 9 pages.

• [16] B. Martinet, Régularisation d’inéquations variationnelles par approximations successives, (French) Rev. Française Informat. Recherche Opérationnelle, 4 (1970), 154–158.

• [17] B. Martinet , Détermination approchée d’un point fixe d’une application pseudo-contractante , Cas de l’application prox, (French) C. R. Acad. Sci. Paris Sér. A-B, 274 (1972), 163–165.

• [18] G. B. Passty , Ergodic convergence to a zero of the sum of monotone operators in Hilbert space, J. Math. Anal. Appl., 72 (1979), 383–390.

• [19] X.-L. Qin, S.-S. Chang, Y. J. Cho , Iterative methods for generalized equilibrium problems and fixed point problems with applications, Nonlinear Anal. Real World Appl., 11 (2010), 2963–2972.

• [20] X.-L. Qin, S. Y. Cho , Convergence analysis of a monotone projection algorithm in reflexive Banach spaces, Acta Math. Sci. Ser. B Engl. Ed., 37 (2017), 488–502.

• [21] X.-L. Qin, J.-C. Yao, Weak convergence of a Mann-like algorithm for nonexpansive and accretive operators , J. Inequal. Appl., 2016 (2016), 9 pages.

• [22] R. T. Rockafellar, Augmented Lagrangians and applications of the proximal point algorithm in convex programming, Math. Oper. Res., 1 (1976), 97–116.

• [23] R. T. Rockafellar , Monotone operators and the proximal point algorithm, SIAM J. Control Optimization, 14 (1976), 877–898.

• [24] D. R. Sahu, J. C. Yao, A generalized hybrid steepest descent method and applications, J. Nonlinear Var. Anal., 1 (2017), 111–126.

• [25] W. Takahashi , Weak and strong convergence theorems for families of nonlinear and nonself mappings in Hilbert spaces, J. Nonlinear Var. Anal., 1 (2017), 1–23.

• [26] S. Takahashi, W. Takahashi, M. Toyoda , Strong convergence theorems for maximal monotone operators with nonlinear mappings in Hilbert spaces, J. Optim. Theory Appl., 147 (2010), 27–41.

• [27] H. K. Xu, Inequalities in Banach spaces with applications, Nonlinear Anal., 16 (1991), 1127–1138.

• [28] S. Yang, Zero theorems of accretive operators in reflexive Banach spaces, J. Nonlinear Funct. Anal., 2013 (2013), 12 pages.


text:
## interference maximum

In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1 and r2 interfere. The parameters for this problem are these: n1 = 1.54, n2 = 1.60, n3 = 1.38, wavelength ? in air = 600 nm. What is the third least thickness that gives an interference maximum of rays r1 and r2?


text:
# JSONDecodeError with backend.run

I'm trying to follow tutorial on pulse simulation described in qiskit.org. The generated PulseQobj is well executed with AER simulator, but it throws JSONDecodeError when using real backends (ex. ibmq-armonk). The PulseQobj passed the validation (validate_qobj_against_schema), so I'm struggling to find the reason. Qobj, generated by QuantumCircuit command, does not raise this error.

I connected to real device with APItoken, and qiskit version is up to date (qiskit 0.22.0). Below is the trial code with real device, and log after backend.run(PulseQobj)

%matplotlib inline

import numpy as np
from qiskit.ignis.characterization.calibrations import rabi_schedules, RabiFitter
from qiskit.pulse import DriveChannel
from qiskit.compiler import assemble
from qiskit.qobj.utils import MeasLevel, MeasReturnType
from qiskit import IBMQ

IBMQ.enable_account(APItoken, proxies = ProxyConfig, verify = certPath)
provider = IBMQ.get_provider(hub=hub, group=group, project=project)
backend = provider.get_backend(device)

# qubit list
qubits = [0]

# drive amplitudes to use
num_exps = 64
drive_amps = np.linspace(0, 1.0, num_exps)

# drive shape parameters
drive_duration = 2048
drive_sigma = 256

# list of drive channels
drive_channels = [DriveChannel(0)]

# construct the schedules
rabi_schedules, xdata = rabi_schedules(amp_list=drive_amps,
qubits=qubits,
pulse_width=drive_duration,
pulse_sigma=drive_sigma,
drives=drive_channels,
inst_map=backend.defaults().instruction_schedule_map,
meas_map=backend.configuration().meas_map)

rabi_qobj = assemble(rabi_schedules,
backend=backend,
meas_level=1,
meas_return='avg',
shots=512)

result = backend.run(rabi_qobj).result()

---------------------------------------------------------------------------
JSONDecodeError                           Traceback (most recent call last)
<ipython-input-17-16f293a93614> in <module>
----> 1 result = backend.run(rabi_qobj).result()

JSONDecodeError: Expecting value: line 1 column 1 (char 0)

• Consider adding the minimum failing example. Also, no need to include the full traceback, just the first and last section would make it... – luciano Oct 16 '20 at 17:16
• Thanks for the comment. I edited the post. – Jin Oct 18 '20 at 23:10

• @Jin I was able to run the example you posted on ibmq_armonk using Qiskit 0.22.0 successfully. Can you try running it without proxies? Some people had reported issues when running with proxies. Also can you post the output of qiskit.__qiskit_version__? Sometimes mismatching versions can happen for whatever reason. – jyu00 Oct 19 '20 at 13:17
• Thanks for the comment. Because I'm behind the institute's firewall, proxy option is required to access IBM account. As you pointed out, in my jupyter environment, qiskit is imported from the wrong path. I resolved this issue, and qiskit.__qiskit_version__ now indicates {'qiskit-terra': '0.16.0', 'qiskit-aer': '0.7.0', 'qiskit-ignis': '0.5.0', 'qiskit-ibmq-provider': '0.11.0', 'qiskit-aqua': '0.8.0', 'qiskit': '0.23.0'}. However, now I got HTTPError: 403 Client Error, IBMQBackendAPIError. So...things are getting complicated. Following error log, it seems API error code 3458. – Jin Oct 20 '20 at 2:50
• Error code 3458 means you've reached the maximum number of concurrent jobs you're allowed. You can use backend.job_limit().maximum_jobs to see what the limit is. And most error codes are documented here: quantum-computing.ibm.com/docs/manage/errors – jyu00 Oct 20 '20 at 12:39
• I've seen wrong error code. It's just HTTPError, RequestsAPIError and IBMQBackendApiError. I think it is still networking error. Meanwhile, as you commented, I checked the backend.job_limit().maximum_jobs and backend.job_limit().active_jobs. Even though there are no active jobs, same error occurs when the job is submitted. – Jin Oct 20 '20 at 23:20


text:
Inaccurate table numbering with LyX

I'm using LyX to create some tables, but the numbering is complete borked.

In the LyX window, the table numbers are accurate. However, when I render to a PDF, they are evens only: 2, 4, 6 etc. (I'm sure I don't need to explain what even numbers are, but you get the point).

Delving into the source code generated:

\selectlanguage{english}%
~\foreignlanguage{british}{}
\begin{table}[H]
\begin{raggedright}
\begin{longtable}[l]{|r|>{\raggedright}m{0.8\textwidth}|c|}
\hline
\selectlanguage{british}%
No.\selectlanguage{english}%
& \selectlanguage{british}%
Requirement\selectlanguage{english}%
& \selectlanguage{british}%
Priority\selectlanguage{english}%
\tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
\end{longtable}
\par\end{raggedright}

\selectlanguage{british}%
\caption{Test table}
\selectlanguage{english}%
\end{table}

This is a test table, hence no data. But, I did notice \begin{table} and \begin{longtable}.

Would I be correct in assuming this is the cause of the problem? If so, does any know how I can go about removing it?

N.b. I've been into the table settings and the long table checkbox is ticked but greyed out, so that's not much use.

-
The syntax should be something like \begin{longtable}[l]{|r|>{\raggedright}m{0.8\textwidth}|c|} \caption{Test table}\\hline(...contents...)\end{longtable}. –  Gonzalo Medina Apr 11 '13 at 19:14

The generated code is full of redundant (or even wrong) things. I don't understand why continuously switching between english and british, for instance: a document should use one English variety, either AmE or BrE.

Using longtable inside a table environment is completely useless. Just a tabular will do. And this is the cause of the wrong numbering, by the way.

The raggedright environment doesn't exist and using the environment form is bound to problems.

\begin{table}
\raggedright
\begin{tabular}[l]{|r|>{\raggedright}m{0.8\textwidth}|c|}
\hline
No. & Requirement & Priority \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
&  & \tabularnewline
\hline
\end{tabular}

\caption{Test table}
\end{table}

-
Thanks for your answer egreg. I guess this is a prime example why using straight latex would be better. –  kyranjamie Apr 15 '13 at 20:41


text:
# Tweening Colors in OpenGL

I'm making a sky gradient in OpenGL by drawing with glColorPointer and glDrawArrays. I would like to be able to change the sky colour from morning to daytime to evening etc. I can either:

• Make a number of sprites and fade them in with my framework
• Somehow tween the color vector in openGL over time and use a single sprite

The second one seems like a more efficient option, but my framework doesn't pass the time delta into the draw method for me to decide how far I've progressed into fading.

Here's my current code:

glDisable(GL_BLEND);
glDisable(GL_DITHER);
glDisable(GL_FOG);
glDisable(GL_LIGHTING);
glDisable(GL_TEXTURE_2D);

CGSize size = [[CCDirector sharedDirector] winSize];
float w = size.width;
float h = size.height;

const GLfloat vertices[] = {
0, 0,
w, 0,
0, h/3,
w, h/3,
0, h*2/3,
w, h*2/3,
0, h,
w, h,
};

const GLubyte colors[] = {
254,255,134,255,
254,255,134,255,
230,157,0,255,
230,157,0,255,
230,60,0,255,
230,60,0,255,
167,0,86,255,
167,0,86,255,
};

glVertexPointer(2, GL_FLOAT, 0, vertices);
glEnableClientState(GL_VERTEX_ARRAY);
glColorPointer(4, GL_UNSIGNED_BYTE, 0, colors);
glEnableClientState(GL_COLOR_ARRAY);

glDrawArrays(GL_TRIANGLE_STRIP, 0, 8);

glDisableClientState(GL_COLOR_ARRAY);
glDisableClientState(GL_VERTEX_ARRAY);

glEnable(GL_TEXTURE_2D);

Which gives me:

• "but my framework doesn't pass the time delta into the draw method" Then your framework is terrible and should be fixed and/or abandoned ASAP. If your drawing code has no way to get a time delta, then that's something that needs to be fixed. Jan 31 '12 at 7:41
• @Nicol I don't think I've ever used a framework that passed dT into the draw method. And you could just as easily encapsulate that functionality in a class in which draw method uses a variable that gets updated in its update method (which presumably has access to dT). Jan 31 '12 at 7:49
• This doesn't really answer your question - but you may get better results by tweening Hue-Saturation-Lightness/Value instead of RGB. It should result in a seemingly more natural transition. Also, nothing is stopping you from calculating the time deltas yourself; or update the sky entity from the update function with the time of day. Jan 31 '12 at 8:22
• @NicolBolas: Excuse me, but why would you need the time delta in the drawing method? Isn't the time delta only interesting in the update-logic, f.e. to know how much to move the ball down the slope? Jan 31 '12 at 8:34
• @Bobby: It depends on what you mean by "the draw method". That is, what the main drawing function's responsibilities are. The OP, for example, wants his drawing function to perform animation on drawing parameters. Is that something he should compute outside of the draw function? It depends on how he wants to structure his code. If there were serious animation, then there would need to be a system to handle animation of all kinds, then feed those values into hooks to the rendering system. But again, it depends on what he wants. Jan 31 '12 at 16:10

I fail to see what's the problem you're having. Update the time of day and tween the colors in the Update method (assuming it has access to the dt) and leave the Draw method only for rendering. I think that's the more logical solution. Also, encapsulate this in some sort of Sky class if you haven't done so already. So I'd definetly stick to the 2nd option. Something like:

class Sky
{
void Update(float dt)
{
// Update timeOfDay based on dt
// Update colors array based on timeOfDay
}

void Draw()
{
// Render sprite using current colors
}

float timeOfDay;
}

Edit

In particular, I think it would be a good idea to start by encapsulating the vertex/color generation code in a method that given a simple array of colors, would return the vertex/color arrays required for calling glVertexPointer and glColorPointer. Then use this method to generate each of the gradient sprites, and finally, given the timeOfDay determine which two of these sprites should be interpolated and by what amount. Then draw them on top of each other and interpolate the alpha value so that there's a cross-fade between them.
• I figured I only need two "GradientFill" sprites and I can load the nextState texture and fade into the current state, then unload the old state and load the next etc. Jan 31 '12 at 12:57


text:
Previous studies of blood cells from atomic bomb survivors have shown that frequencies of chromosome aberrations and somatic mutations are elevated in heavily exposed survivors and that T-cell functions and the number of mature T cells are decreased in the survivors who were exposed to radiation as adults. Current progress in flow cytometry allows a sophisticated analysis of various subsets of T, B and NK cells. In the present study, proportions of such subsets in peripheral blood lymphocytes from atomic bomb survivors (159 survivors estimated to be exposed to ≥1.5 Gy) and 252 controls were measured using multiple combinations of monoclonal antibodies to lymphocyte differentiation antigens to investigate whether the previous radiation exposure had altered the composition of the subsets. Among T-cell subsets, the proportion of CD4+ T-cell subsets was decreased significantly in the heavily exposed survivors; this tendency was apparent for the <tex-math>${\rm CD}4^{+}{\rm CD}45{\rm RA}^{+}$</tex-math> naive T-cell subset. However, there were no significant differences in the proportions of CD8+ T-cell subsets between the exposed survivors and controls. As for the B-cell subsets, the proportion of both CD5+ and <tex-math>${\rm CD}5^{-}$</tex-math> B cells as well as <tex-math>${\rm CD}23^{+}$</tex-math> and <tex-math>${\rm CD}23^{-}$</tex-math> B cells increased in the heavily exposed survivors. Further, no effect of radiation was found in the proportion of NK-cell subsets. These results strongly suggest that previous radiation exposure altered the composition of T and B cells in the peripheral blood of atomic bomb survivors, and they raise the possibility that atomic bomb radiation may have affected the developmental processes of T and B cells.

This content is only available as a PDF.


text:
# Find the Surface Area sphere (3.5)

r=3.5
The surface area of a sphere is equal to 4 times Pi π times the radius squared.
Substitute the value of the radius r=3.5 into the formula to find the surface area of the sphere. Pi π is approximately equal to 3.14.
4⋅π⋅3.52
Raise 3.5 to the power of 2.
4π⋅12.25
Multiply 4π⋅12.25.
Multiply 12.25 by 4.
49π
Multiply 49 by π.
153.93804002
153.93804002
Find the Surface Area sphere (3.5)

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text:
## how to evaluate integrals with bounds

Step 2: Integrate the function using the usual rules of integration. If one or both integration bounds a and b are not numeric, int assumes that a <= b unless you explicitly specify otherwise. The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral, D f (x,y) dA D f ( x, y) d A. where D D is any region. fg dx = fg f gdx. 1 3 6x25x +2dx 3 1 6 x 2 5 x + 2 d x. Step 2: Identify the calculus limits of the integral. A Definite Integral has start and end values: in other words there is an interval [a, b]. In these cases, the interval of integration is said to be over an infinite interval. Here R.H.S. Additionally, you can access the integration template from the Functions menu on the keyboard, under Miscellaneous functions. Integral calculator is an online tool that calculates the antiderivative of a function. Step 1: Press the F3 button: Step 2: Press "2": Step 3: Type in your function. Let's take a look at an example that will also show us how we are going to deal with these integrals. u = f(x) v = g(x) du = f (x)dx dv = g (x)dx. The definite integral of on the interval is most generally defined to be. You can do this for any math operator. Hence, it can be said F is the anti-derivative of f. Definite integrals are also known as Riemann . The limits of integration as dxdydxdy order are required to determine the limits of integration for the equivalent integral dydxdydx order. Apr 20 Integrals with Functions as Bounds. Example 2 Evaluate each of the following. Example 1: Evaluate Definite Integral limits using integration by parts with both Rule 1 and 2. 2 Evaluate a triple integral by expressing it as an iterated integral NZ Grapher was designed for New Zealand Schools by a New Zealand Teacher . It keeps running and running and is never able to resolve the equation. Step-by-Step Examples. Here R.H.S. Integration Worksheet Introduction to triple integral finding bounds Funcions 3D plotter calculates the analytic and numerical integral and too calculates partial derivatives with respect to x and y . Read more. If the upper limit and the lower limit of the independent variable of the given function or integrand is specified, its integration is expressed using definite integrals. . For this example problem, use "b" to replace the upper infinity symbol. For convenience of computation, a special case of the above definition uses subintervals of equal length and sampling points chosen to be the right-hand endpoints of the subintervals. The integral of 1 x2 is -1 x, so: Step 3: Evaluate the definite integral: As b approaches infinity, -1/b tends towards zero. . Evaluate the triple integral if E E E is the region below z = x + y 1 z=x+y-1 z = x + y 1 but above the . The fundamental theorem of calculus is a theorem that connects the concept of differentiation with the co. To evaluate the integral in G, use the release function to ignore the 'Hold' option.

To evaluate the integral in G, use the release function to ignore the 'Hold' option. To change the function and limits of integration from rectangular coordinates to polar coordinates, we'll use the conversion formulas. The same integral reduced to an iterated integral with bounds depending on the previous . I have been having some trouble getting the sympy module to evaluate a definite integral. Solution: For solving the above definite integral problem with integration by parts using Rule 1, we have to apply limits after the end of our result. Where, a and b are the lower and upper limits. 0 4 t(t2) dt 4 0 t ( t 2) d t. 2 1 2w5 w+ . There are two parts of the Fundamental Theorem of Calculus: Part One. b a f ( x) d x = F ( a) F ( b) where F (x) is the antiderivative of f (x) a b f ( x) d x = F ( a) F ( b) where F (x) is the antiderivative of f (x . Example 1 Evaluate the following integral. Apr 20 Integrals with Functions as Bounds. Search: Improper Integral Calculator. Example question: Evaluate: To put this in English, the question is asking you to find the definite integral of f (x) = x 2 on the closed interval [1, 3] (1, 3 are called the bounds of integration ). These values are typically denoted at the top and bottom of the integral sign. When evaluating limits, we want to first check to see if the function is continuous When evaluating limits, we want to first check to see if the function is continuous. Calculus Examples. Integrals. Example 1: Evaluate Definite Integral limits using integration by parts with both Rule 1 and 2.

. 1 3 6x25x +2dx 3 1 6 x 2 5 x + 2 d x. Step 2: Identify the calculus limits of the integral. Step-by-Step Examples. Indefinite integrals are functions while definite integrals are numbers. Step 1: Replace the infinity symbol with a finite number.

Step 1: Press the F3 button: Step 2: Press "2": Step 3: Type in your function. The curve on this page (250/ (25+x^2)) looks like it should be at least twice as large as that under the curve of 1/x. 3 4 0 x2dx 3 0 4 x 2 d x. If the upper limit and the lower limit of the independent variable of the given function or integrand is specified, its integration is expressed using definite integrals. command brings the integrandum closer to the integral sign. a and b (called limits, bounds or boundaries) are put at the bottom and top of the "S", like this: Definite Integral (from a to b) Indefinite Integral (Opens a modal) Definite integral of rational function. Hence, it can be said F is the anti-derivative of f. Definite integrals are also known as Riemann . We'll allow the upper bound to be 2 while the lower bound is 1. MathJax TeX Test Page. The notation for the general triple integrals is, $\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}}$ Let's start simple by integrating over the box, When finding a definite integral using integration by parts, we should first find the antiderivative (as we do with indefinite integrals), but then we should also evaluate the antiderivative at the boundaries and subtract. Do not use waypoints to specify singularities. Remember also that when you convert d A dA d A or d y d x dy\ dx . Improper integral is definite integral, which is unlimited or expanded function, or the region Definite Integral Calculator Compute the improper integral R 1 a 1 x dx Calculus: Integral with adjustable bounds A tutorial on the definition of definite integrals, properties of definite integrals, relationship between definite integrals and areas . Integral calculator is a mathematical tool which makes it easy to evaluate the integrals. Line integration calculator shows you all of the steps required to evaluate the integrals. Example #1: Definite Integral on the TI 89. A definite integral is denoted as: F ( a) - F ( b) = a b f ( x) d x. Let's work some more examples. . It can also evaluate integrals that involve exponential, logarithmic, trigonometric, and inverse trigonometric functions, so long as the result comes out in terms of the same set of functions. Finally, note the difference between indefinite and definite integrals. In this kind of integral one or both of the limits of integration are infinity. f g d x = f g f g d x. Since 3 3 is constant with respect to x x, move 3 3 out of the integral. In this case, all of the integration limits and waypoints must be finite. The integral, also called antiderivative, of a function, is the reverse process of differentiat. As you can see, adding the macro \limits to your code makes the integral look good. Finally, note the difference between indefinite and definite integrals. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Share. By the Power Rule, the integral of x2 x 2 with respect to x x is 1 3x3 1 3 x 3. These values are typically denoted at the top and bottom of the integral sign. Now let's do a triple integral without limits of integration to see how it's different. Example 2 Evaluate each of the following. Browse other questions tagged calculus integration definite-integrals indefinite-integrals or ask your own question. 1 1 x2 dx 1 1 x 2 d x. The Overflow Blog Celebrating the Stack Exchange sites that turned ten years old in Spring 2022 Integration by parts: definite integrals. Example #1: Definite Integral on the TI 89. Example question: Evaluate: To put this in English, the question is asking you to find the definite integral of f (x) = x 2 on the closed interval [1, 3] (1, 3 are called the bounds of integration ). David Witten. F (x) is the integral of f (x), and if f (x) is differentiated, F (x) is obtained. 4 0 3x2dx 0 4 3 x 2 d x. The Integral Calculator lets you calculate integrals and antiderivatives of functions online for free! After the Integral Symbol we put the function we want to find the integral of (called the Integrand). In these cases, the interval of integration is said to be over an infinite interval. But the very next video "Divergent Improper Integral" shows an area of infinity under the curve of 1/x. Definite Integrals.

Simply type int in an expression line to bring up an integration template. Indefinite integrals are functions while definite integrals are numbers. This is not the easiest formula to use however. Now at first this might seem daunting, I have this rational expression, I have xs in the numerators and xs in the denominators, but we just have to remember, we just have to do some algebraic manipulation, and this is going to seem a lot more attractable. Example 1 Evaluate the following integral. Evaluate the integral from x=0 to x=Inf, adjusting the absolute and relative tolerances. The upper bound is the value up top and the lower bound is the value at the bottom of the symbol. Integral Calculator. We always work our way "inside out" in order to evaluate iterated integrals. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. By the Power Rule, the integral of x2 x 2 with respect to x x is 1 3x3 1 3 x 3. Fundamental Theorem of Calculus. (Opens a modal) Intuition for second part of fundamental theorem of calculus. Solution: For solving the above definite integral problem with integration by parts using Rule 1, we have to apply limits after the end of our result. The Mean Value Theorem for Definite Integrals: If f ( x) is continuous on the closed interval [ a, b ], then at least one number c exists in the open interval ( a, b) such that. When you use . It helps you practice by showing you the full working (step by step integration). Follow Evaluate the Integral. The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral, D f (x,y) dA D f ( x, y) d A. where D D is any region. Evaluate the Integral.

explain the meaning of the terms 'proper fraction' and 'improper fraction' express an algebraic fraction as the sum of its partial fractions Contents 1 Series estimation with integrals The Voovers Integral Calculator will get you a reliable solution extremely quickly are improper integrals in the sense of the above definitions integrate(f, x . We'll allow the upper bound to be 2 while the lower bound is 1. Improve this answer. Example. Our calculator allows you to check your solutions to calculus exercises.

Definite integrals of a function f (x) from a to b when the function f is continuous in the closed interval [a, b]. This calculator can help managing that and converting the units of Lambda and Mu to other ones Since the triangle above the x-axis has positive signed area and the triangle below x-axis has negative signed area, the total area represented by the integral is: Human-computer interaction (HCI) is the study of designing, implementing, and evaluating the interactive interfaces used by humans HCI is . Rule 2: The second rule said that you can use the limit as you are going with integrals of . If one or both integration bounds a and b are not numeric, int assumes that a <= b unless you explicitly specify otherwise. In the previous section we looked at double integrals over rectangular regions. The fundamental theorem of calculus and definite integrals. (Opens a modal) Area between a curve and the x-axis. The integral, also called antiderivative, of a function, is the reverse process of differentiat. 3 4 0 x2dx 3 0 4 x 2 d x.

Equation When I try to run the following code the program fails to finish. Search results for "triple integrals" Triple Integral Calculator 3) > restart: Setting limits of integration and evaluating NZGrapher is a web based graphing tool Graphing Calculator Polar Curves Derivative Calculator Integral Calculator Formulas and Notes Equation If we program a grapher to do the work of finding the points, Eulers Method can . - [Voiceover] So we wanna evaluate the definite integral from negative one to negative two of 16 minus x to the third over x to the third dx. 0 4 t(t2) dt 4 0 t ( t 2) d t. 2 1 2w5 w+ . . Step 3: Perform the integration of the function . Definite integrals of a function f (x) from a to b when the function f is continuous in the closed interval [a, b]. David Witten. Thus, each subinterval has length. of the equation means integral of f (x) with respect to x. f (x)is called the integrand.


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# m03 q 115

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m03 q 115 [#permalink]  12 Jan 2012, 05:36
If two lines have slopes m and n, respectively, are they perpendicular?

1. m*n = -1
2. m = -n

Wouldn't the answer be D? Each alone is sufficient?

1. Sufficient - we know that whenever slopes are negative reciprocals of each other, the lines are perpendicular (OK).
2. Sufficient - we know that the slopes are NOT negative reciprocals.. wouldn't this mean that we KNOW for a fact that the lines are NOT perpendicular?

The answer given is A, what am I missing. Can lines whose slopes are just reciprocals be perpendicular?

Thanks!
M
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Re: m03 q 115 [#permalink]  12 Jan 2012, 05:46
Expert's post
getzonator wrote:
If two lines have slopes m and n, respectively, are they perpendicular?

1. m*n = -1
2. m = -n

Wouldn't the answer be D? Each alone is sufficient?

1. Sufficient - we know that whenever slopes are negative reciprocals of each other, the lines are perpendicular (OK).
2. Sufficient - we know that the slopes are NOT negative reciprocals.. wouldn't this mean that we KNOW for a fact that the lines are NOT perpendicular?

The answer given is A, what am I missing. Can lines whose slopes are just reciprocals be perpendicular?

Thanks!
M

For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is $$m$$ then the line prependicular to it will have the slope $$-\frac{1}{m}$$. In other words, the two lines are perpendicular if and only the product of their slopes is -1.

For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

If two lines have slopes m and n, respectively, are they perpendicular?

(1) m*n = -1 --> directly gives an answer YES to the question.

(2) m = -n --> now, if for example m=3=-(-3)=-n then the lines are not perpendicular but if m=1=-(-1)=-n, so if m=1 and n=-1 (or m=-1=-1=-n) then as mn would be equal to -1 the lines would be perpendicular. Not sufficient.

getzonator you should have spotted that there was something wrong with your solution as on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

So we can not have answer YES from statement (1) and answer NO from statement (2), as in this case statements would contradict each other.

Hope it helps.
_________________
Re: m03 q 115   [#permalink] 12 Jan 2012, 05:46
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01p8418q67p
 Title: TERM LOGIC, MODERN PREDICATE LOGIC AND LOGICAL PLURALISM Authors: Furchtgott, Godfrey Advisors: Halvorson, Hans Contributors: Boris, Kment Department: Philosophy Class Year: 2016 Abstract: In this paper, I explicate Fred Sommers' natural-language based Term Functor Logic and compare it with the canonical Modern Predicate Logic. I suggest that the processes of regimentation as well as the formal underpinnings of the logic systems are important to consider in comparing the systems. I argue that in the absence of absolute inferential differences between the systems, we ought to adopt a logically pluralist position. Extent: 38 pages URI: http://arks.princeton.edu/ark:/88435/dsp01p8418q67p Type of Material: Princeton University Senior Theses Language: en_US Appears in Collections: Philosophy, 1924-2016

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# Inertial mass vs gravitational mass

Source: Introduction to Mechanics Kleppner and Kolenkow:

The “mass” in the law of gravitation (gravitational mass) measures the strength of gravitational interaction and is operationally distinct from the “mass” (inertial mass) that characterizes inertia in Newton’s second law. Why gravitational mass is proportional to inertial mass is a deep mystery. Newton recognized the mystery and confirmed the fact experimentally to an accuracy of about 1% by observing that the period of a pendulum does not depend on the material of the pendulum.

What experiment is the author talking about? How did Newton find the relation between inertial mass and gravitational mass by knowing that period of pendulum doesn't depends upon material of pendulum? Couldn't just catch it.

For small oscillations, you find that $$T=2\pi\sqrt{\frac{L}{g}}$$, where $L$ is the string lenght, $g$ the gravity, and $T$ the period. You see that this doesn't depend on the mass.

But the key is the process of obtaining this. There are two forces: tension and weight:

The tension must cancel out with the weight's component in that direction. The remaaining "horizontal" component of weight is the responsible of the oscillation.

This component of the weight: $m g \sin(\theta)$ is the only force (the other ones cancel each other).

So, using Newton's law, you'd get

$$m_I\cdot a = m_g\cdot g\cdot \sin(\theta)$$

Notice that "sum of forces = inertial mass times acceleration", so the left hand side has inertial mass $m_I$. However, the weight is given by "gravitational mass times gravity" $m_g\cdot g$.Then it's multipleid by the sine because we're taking its horizontal component.

If oscillations are small enough, we can make an approximation: $sin(\theta)\approx \theta$. This approximaton makes the problem much easier to solve, and it is very accurate.

So we have $$m_I\cdot a \approx m_g\cdot g \cdot \theta$$

If you replace $a=l \alpha$, where $\alpha$ is the angular acceleration, then we have

$$m_I\cdot l \cdot \alpha \approx m_g\cdot g \cdot \theta$$

which you can rearrange as

$$\alpha\approx \frac{m_g}{m_I} \frac{g}{l} \theta$$

or equivalently as

$$\ddot{\theta}- \frac{m_g}{m_I} \frac{g}{l} \theta \approx 0$$

since $\alpha=\ddot{\theta}$. This is the famous equation of the harmonic oscilaltor, with a frequency given by what is next to the angle:

$$\omega^2 = \frac{m_g}{m_I} \frac{g}{l}$$

Which implies that the period is

$$T=\frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m_I}{m_g} \frac{L}{g}}$$

The thing is that, experimentally, $m_I=m_g$ is found, and hence the period depends only on the lenght, and not on the gravitational mass we add on it.

• This shows inertial mass equals passive gravitational mass, which is not so mysterious now after we have GR. But why active gravitational mass equals passive gravitational mass equals inertial mass? – velut luna Mar 25 '18 at 10:09
• How was it experimentally proven that, mI=mg? Eötvös experiment? Was it done at the time of Newton? – suiz Mar 25 '18 at 10:30
• I've got the answer from the link mathpages.com/home/kmath582/kmath582.htm . – suiz Mar 28 '18 at 6:28