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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: Find the value of $k$ for which the roots of the quadratic equation
$(k - 5)x^2 + 2(k - 5) x + 2 = 0$ are equal.

Given equation: $(k - 5)x^2 + 2(k - 5) x + 2 = 0$

For roots to be equal, $b^2 - 4ac = 0$

$\Rightarrow [ 2(k-5)]^2 - 4 (k-5)(2) = 0$

$\Rightarrow k^2 +25-10k-2k +10 = 0$

$\Rightarrow k^2 - 12k + 35 = 0$

$\Rightarrow (k-7)(k-5) = 0$

$\Rightarrow k = 7$ or $k = 5$

$\\$

Question 2: Find the value of $y$ for which the distance between the points $(2, -3)$ and $(10, y)$ is $10$ units.

Given distance between the points $(2, -3)$ and $(10, y)$ is $10$ units.

$\Rightarrow \sqrt{(10-2)^2 + (y+3)^2} = 10$

$\Rightarrow 8^2 + (y+3)^2 = 10^2$

$\Rightarrow y^2 + 9 + 6y = 36$

$\Rightarrow y^2 + 6y - 27 = 0$

$\Rightarrow y^2 + 9y - 3y - 27 = 0$

$\Rightarrow y(y+9) -3(y+9) = 0$

$\Rightarrow (y+9) (y-3) = 0$

$\Rightarrow y = -9$ or $y = 3$

$\\$

Question 3: Write whether the rational number $\frac{13}{3125}$ has a decimal expansion which is terminating or non-terminating repeating.

$\frac{13}{3125}$ $=$ $\frac{13}{2^0 \times 5^5}$ $=$ $\frac{13}{2^0 \times 5^5} \times \frac{2^5}{2^5}$ $=$ $\frac{13 \times 2^5}{10^5}$ $=$ $\frac{416}{10^5}$ $=$ $0.00416$

$\\$

Question 4: Write the $m^{th}$ term of the A.P. $\frac{1}{k}$, $\frac{1+k}{k}$, $\frac{1+2k}{k}$, $\cdots$

From the given AP

$a =$ $\frac{1}{k}$ and $d =$ $\frac{1+k}{k}$ $-$ $\frac{1}{k}$ $=$ $\frac{k}{k}$ $= 1$

$\therefore T_m = a+(m-1)d$

$\Rightarrow T_m =$ $\frac{1}{k}$ $+ (m-1)$

$\Rightarrow T_m =$ $\frac{k(m-1)+1}{k}$

$\\$

Question 5: If $\sin \theta + \cos \theta = 2 \cos (90^o - \theta)$, find the value of $\cot \theta$

$\sin \theta + \cos \theta = 2 \cos (90^o - \theta)$

$\sin \theta + \cos \theta = 2 \sin \theta$

$\Rightarrow 1 + \cot \theta = \sqrt{2}$

$\Rightarrow \cot \theta = \sqrt{2} -1$

$\\$

Question 6: $DE$ is drawn parallel to the base $BC$ of a $\triangle ABC$, meeting $AB$ at $D$ and $AC$ at $E$. If  $\frac{AB}{BD}$ $= 4$ and $CE = 2$ cm, find $AE$.

Given: $DE \parallel BC,$ $\frac{AB}{BD}$ $= 4$

$\Rightarrow \triangle ABC \sim \triangle ADE$ by AAA criterion

$\frac{AB}{AD}$ $=$ $\frac{AC}{AE}$

$\frac{AB}{AB-BD}$ $=$ $\frac{AC}{AC-CE}$

$\frac{1}{1-\frac{BD}{AB} }$ $=$ $\frac{1}{1- \frac{CE}{AC}}$

$1-$ $\frac{CE}{AC}$ $= 1-$ $\frac{BD}{AB}$

$\frac{CE}{AC}$ $=$ $\frac{BD}{AB}$

$AC =$ $\frac{AB}{BD}$ $\times CE = 4 \times 2 = 8$ cm

$\therefore AE = AC = CE = 8 - 2 = 6$ cm

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains $5$ red balls and some blue balls. If the probability of drawing a blue ball from the bag is three times that of a red ball, find the number of blue balls in the bag.

Number of Red balls $= 5$

Let the number of Blue balls $= x$

$\therefore \frac{x}{5+x}$ $= 3 \times$ $\frac{5}{5+x}$

$\Rightarrow x = 15$

No of Blue balls in the bag $= 15$

$\\$

Question 8: The $5^{th}$ and $15^{th}$ terms of an A.P. are $13$ and $-17$ respectively. Find the sum of first $21$ terms of the A.P.

Let the first term $= a$ and the common difference $= d$

$T_5 = a + (5-1)d = 13$

$\Rightarrow a + 4d = 13$    … … … … … i)

$T_{15} = a + (15-1)d = -17$

$\Rightarrow a + 14d = -17$    … … … … … ii)

Subtracting ii) from i)

$10d = - 30 \Rightarrow d = -3$

$\therefore a = 13 - 4(-3) = 25$

$\therefore a = 25$ and $d = -3$

We know $S_n =$ $\frac{n}{2}$ $[2a + (n-1)d ]$

$\therefore S_{21} =$ $\frac{21}{2}$ $[2(25) + (21-1)(-3) ]$

$\Rightarrow S_{21} =$ $\frac{21}{2}$ $[50) + 20(-3) ]$

$\Rightarrow S_{21} =$ $\frac{21}{2}$ $[50 -60 ]$

$\Rightarrow S_{21} = 21 \times (-5) = -105$

Therefore the sum of the first $21$ terms is $- 105$

$\\$

Question 9: Using Euclid’s Division Algorithm, find the HCF of $255$ and $867$.

$255 ) \overline{867} ( 3 \\ \hspace*{0.7cm} \underline{765} \\ \hspace*{0.8cm} 102 ) \overline{255} ( 2 \\ \hspace*{1.5cm} \underline{204} \\ \hspace*{1.7cm} 51 ) \overline{102} ( 2 \\ \hspace*{2.25cm} \underline{102} \\ \hspace*{2.5cm} \times$

According to Eculid’s division theorem, any positive number can be expressed as $a = bq+r$ where $q$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

$\therefore 867 = 51 \times 17 + 0$

So HCF of $867$ and $255$ is $51$

$\\$

Question 10: If the point $(0, 2)$ is equidistant from the points $(3, k)$ and $(k, 5)$, find the value of $k$.

Given point $(0, 2)$ is equidistant from the points $(3, k)$ and $(k, 5)$

Applying distance formula

$\sqrt{(3-0)^2+(k-2)^2} = \sqrt{(k-0)^2+(5-2)^2}$

$\Rightarrow 9 + (k-2)^2 = k^2 + 9$

$\Rightarrow k^2 + 4 - 4k = k^2$

$\Rightarrow k = 1$

$\\$

Question 11: Find the value of $'a'$ for which the pair of linear equations $2x + 3y = 7$ and $4x + ay = 14$ has infinitely many solutions.

If the system of equations are $ax_1 + b_1y + c_1 = 0$ and $ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:
$\frac{a_1}{a_2}$ $=$ $\frac{b_1}{b_2}$ $=$ $\frac{c_1}{c_2}$

$\frac{2}{4}$ $=$ $\frac{3}{a}$ $=$ $\frac{7}{14}$

From first two terms $a = 6$

From Last two terms $a = 6$

Therefore for $a = 6$ equations $2x + 3y = 7$ and $4x + ay = 14$ has infinitely many solutions.

$\\$

Question 12: A card is drawn at random from a well shuffled pack of $52$ playing cards. Find the probability of getting (i) a red king (ii) a queen or a jack

Total number of cards $= 52$

Number of Red kings $= 2$

Number of queens and jacks $= 8$

i) Probability (Red King) $=$ $\frac{2}{52}$ $=$ $\frac{1}{26}$

ii) Probability (Queen or Jack) $=$ $\frac{8}{52}$ $=$ $\frac{2}{13}$

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Show that any positive odd integer is of the form $4q + 1$ or $4q + 3$ for some integer $q$.

Let $a$ be any positive integer.

Eculid’s division theorem, any positive number can be expressed as $a = bq+r$ where $q$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

Take $b = 4 \Rightarrow a = 4q + r$

Since $0 \leq r < 4$, the possible remainders are $0, 1, 2, 3$

That is $a$ can be $4q, 4q+1, 4q+2$ or $4q+3$

Since $a$ is odd, $a$ cannot be $4q$ or $4q+2$

Therefore any odd integer is of the form $4q+1$ or $4q+3$.

$\\$

Question 14: The ten’s digit of a number is twice its unit’s digit. The number obtained by interchanging the digits is $36$ less than the original number. Find the original number.

Let tens digit be $x$ and unit digit be $y$

Therefore the number $=10x+y$

Interchanged number $=10 y+x$

And (tens digit is twice the unit ) That is, $x=2y$    … … … … … i)

Now according to the question

$10 y+x=10x+y-36$

$\Rightarrow 9y =9x-36$

$\Rightarrow y=x-4$  from i)

$\Rightarrow y=2y-4$

$\Rightarrow y=4$

Then $\Rightarrow x=2y=2 (4)=8$

Therefore the number is $84$

$\\$

Question 15: The line segment joining the points $A(2, 1)$ and $B(5, -8)$ is trisected at the points $P$ and $Q$, where $P$ is nearer to $A$. If $P$ lies on the line $2x - y + k = 0$, find the value of $k$.

Or

The $x-$ coordinate of a point $P$ is twice its $y-$ coordinate. If $P$ is equidistant from the points $Q(2, -5)$ and $R(-3, 6)$, find the coordinates of $P$.

$AP : AB = 1:3$ (trisects)

$\frac{AP}{AB}$ $=$ $\frac{1}{3}$

$\Rightarrow \frac{AP}{AP+PB}$ $=$ $\frac{1}{3}$

$\Rightarrow 3AP = AP + PB$

$\Rightarrow 2AP = PB$

$\Rightarrow \frac{AP}{PB}$ $=$ $\frac{1}{2}$

$\Rightarrow AP : PB = 1:2$

Applying section formula

Coordinates of $P = \Big($ $\frac{1 \times 5 + 2 \times 2}{1 + 2}$ $,$ $\frac{1 \times (-8) + 2 \times (1)}{1 + 2}$ $\Big)$

$\Rightarrow P = \Big($ $\frac{5+4}{3}$ $,$ $\frac{-8+2}{3}$ $\Big)$

$\Rightarrow P = (3, -2)$

Since $P(3, -2)$ lies on $2x - y + k = 0$

$\therefore 2(3) - (-2) + k = 0$

$\Rightarrow 6 + 2 + k = 0$

$\Rightarrow k = -8$

Or

Let $P(x, y)$ be the required point

Given: $P$ is equidistant from the points $Q(2, -5)$ and $R(-3, 6)$

$\therefore PQ = PR \Rightarrow PQ^2 = PR^2$

$\Rightarrow (x-2)^2 + (y + 5)^2 = (x+3)^2 + (y-6)^2$

$\Rightarrow x^2 + 4 - 4x + y^2 + 25 + 10y = x^2 + 9 + 6x + y^2 + 36 - 12y$

$\Rightarrow -10x + 22y - 16 = 0$

Now $x = 2y$

$\Rightarrow -10(2y) + 22y = 16$

$\Rightarrow -20y + 22y = 16$

$\Rightarrow 2y = 18$

$\Rightarrow y = 8$

$\therefore x = 2 \times 8 = 16$

Hence $P$ is $(16, 8)$

$\\$

Question 16: Show that $1$, $\frac{1}{2}$ and $-2$ are the zeroes of the polynomial $2x^3 + x^2 - 5x + 2$.

Comparing given cubic equation to $ax^3 + bx^2 + cx + d$ we get $a = 2, b = 1, c = -5$ and $d = 2$

Take $\alpha =$ $\frac{1}{2}$ $, \ \ \beta = 1$ and $\gamma = -2$

Now we verify the relations between zeros and their coefficients

$\alpha + \beta + \gamma =$ $\frac{1}{2}$ $+ 1 - 2 =$ $\frac{-1}{2}$ $=$ $\frac{-b}{a}$

$\alpha . \beta + \beta . \gamma + \gamma . \alpha =$ $\frac{1}{2}$ $\times 1 + 1 \times (-2) + (-2) \times$ $\frac{1}{2}$ $=$ $\frac{1}{2}$ $-2 - 1 =$ $\frac{-5}{2}$ $=$ $\frac{c}{a}$

$\alpha . \beta . \gamma =$ $\frac{1}{2}$ $\times 1 \times (-2) = -1 =$ $\frac{-2}{2}$ $=$ $\frac{-d}{a}$

Hence verified.

Question 17: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Given: Circle with center $O$. Let $PA$ and $PB$ be tangents from external point P.

To Prove: $\angle APB + \angle AOB = 180^o$

Proof: Since $PA$ is a tangent, $OA \perp PA \Rightarrow \angle OAP = 90^o$

Similarly, since $PB$ is a tangent, $OB \perp PB \Rightarrow \angle OBP = 90^o$

In Quadrilateral $OAPB$

$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^o$

$\Rightarrow 90^o + \angle APB + 90^o + \angle AOB = 360^o$

$\Rightarrow \angle APB + \angle AOB = 180^o$

Hence proved.

$\\$

Question 18: $S$ and $T$ are points on the sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.

Or

In an equilateral $\triangle ABC, D$ is a point on the side $BC$ such that $BD = \frac{1}{3} BC$. Prove that $9AD^2 = 7AB^2$.

Given: $\triangle PQR$  and points $S$ and $T$ on sides of $PR$ and $QR$

$\angle P = \angle RTS$

To Prove: $\triangle RPQ \sim \triangle RTS$

Proof: In $\triangle RPQ$ and $\triangle RTS$

$\angle P = \angle RTS$ (given)

$\angle PRQ = \angle TRS = \angle R$ (common angle)

$\therefore \triangle RPQ \cong \triangle RTS$  (by AA criterion)

Hence proved.

Or

Given: $\triangle ABC$ is an equilateral triangle.

$\Rightarrow AB = BC = CA$

Also $BD =$ $\frac{1}{3}$ $BC$

To Prove: $9 AD^2 = 7 AB^2$

Construction: Draw $AE \perp BC$

Proof: Consider $\triangle AEB$ and $\triangle AEC$

$AE$ is common

$\angle AEB = \angle AEC = 90^o$

$AB = AC$ (equilateral triangle)

$\triangle AEB \cong \triangle AEC$ (By RHS criterion)

$\therefore BE = EC$

$\Rightarrow BE =$ $\frac{BC}{2}$

$\Rightarrow BD + DE =$ $\frac{BC}{2}$

$\Rightarrow$ $\frac{BC}{3} + DE =$ $\frac{BC}{2}$

$\Rightarrow DE =$ $\frac{BD}{6}$

Using Pythagoras theorem,

In $\triangle AEB$

$AB^2 = AE^2 + BE^2 \Rightarrow AE^2 = AB^2 - BE^2$   … … … … … i)

In $\triangle AED$

$AD^2 = AE^2 + DE^2 \Rightarrow AE^2 = AD^2 - DE^2$   … … … … … ii)

From i) and ii)

$AB^2 - BE^2 = AD^2 - DE^2$

$\Rightarrow AB^2 - \Big($ $\frac{BC}{2}$ $\Big)^2 = AD^2 - \Big($ $\frac{BC}{6}$ $\Big)^2$

$\Rightarrow AB^2 -$ $\frac{BC^2}{4}$ $= AD^2 -$ $\frac{BC^2}{36}$

$\Rightarrow AB^2 = AD^2 +$ $\frac{8}{36}$ $BC^2$

$AB^2 = AD^2 +$ $\frac{2}{9}$ $BC^2$

But $BC = AB$

$\therefore AB^2 -$ $\frac{2}{9}$ $AB^2 = AD^2$

$\Rightarrow 7AB^2 = 9AD^2$

Hence proved.

$\\$

Question 19:  Prove that: $\frac{1}{\mathrm{cosec} \theta + \cot \theta}$ $-$ $\frac{1}{\sin \theta}$ $=$ $\frac{1}{\sin \theta} - \frac{1}{\mathrm{cosec} \theta - \cot \theta}$

Or

If $\tan \theta + \sin \theta = m, \tan \theta - \sin \theta = n$, show that $m^2 - n^2 = 4 \sqrt{mn}$

Prove $\frac{1}{\mathrm{cosec} \theta + \cot \theta}$ $-$ $\frac{1}{\sin \theta}$ $=$ $\frac{1}{\sin \theta}$ $-$ $\frac{1}{\mathrm{cosec} \theta - \cot \theta}$

or Prove $\frac{1}{\mathrm{cosec} \theta + \cot \theta}$ $+$ $\frac{1}{\mathrm{cosec} \theta - \cot \theta}$ $=$ $\frac{2}{\sin \theta}$

$\Rightarrow$ $\frac{\mathrm{cosec} \theta - \cot \theta + \mathrm{cosec} \theta + \cot \theta}{\mathrm{cosec}^2 \theta - \cot^2 \theta}$ $=$ $\frac{2}{\sin \theta}$

$\Rightarrow$ $\frac{2 \mathrm{cosec} \theta}{\mathrm{cosec}^2 \theta - \cot^2 \theta}$ $=$ $\frac{2}{\sin \theta}$

$\Rightarrow$ $\frac{2}{ \sin \theta \Big( \frac{1}{\sin^2 \theta} - \frac{ \cos^2 \theta }{ \sin^2 \theta } \Big) }$ $=$ $\frac{2}{ \sin \theta }$

$\Rightarrow$ $\frac{2\sin^2 \theta }{\sin \theta (1 - \cos^2 \theta )}$ $=$ $\frac{2}{ \sin \theta }$

$\Rightarrow$ $\frac{2}{ \sin \theta }$  $\frac{ \sin^2 \theta }{ \sin^2 \theta }$ $=$ $\frac{2}{ \sin \theta }$

$\Rightarrow$ $\frac{2}{ \sin \theta }$ $=$ $\frac{2}{ \sin \theta }$

Hence proved.

Or

Given:

$\tan \theta + \sin \theta = m$

$\tan \theta - \sin \theta = n$

$\therefore m^2 - n^2 = ( \tan \theta + \sin \theta )^2 - ( \tan \theta - \sin \theta )^2$

$= \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta - \tan^2 \theta - \sin^2 \theta + 2 \tan \theta \sin \theta$

$= 4 \tan \theta \ \sin \theta$

$4 \sqrt{mn} = 4 \sqrt{( \tan \theta + \sin \theta)( \tan \theta - \sin \theta)}$

$= 4 \sqrt{\tan^2 \theta - \sin^2 \theta }$

$= 4 \sqrt{\frac{ \sin^2 \theta }{ \cos^2 \theta } - \sin^2 \theta }$

$= 4 \sqrt{ \sin^2 \theta \Big( \frac{1}{ \cos^2 \theta} - 1 \Big) }$

$= 4 \sqrt{\sin^2 \theta \frac{1 - \cos^2 \theta }{\cos^2 \theta } }$

$= 4$ $\sqrt{ \frac{ \sin^2 \theta . \sin^2 \theta }{ \cos^2 \theta } }$

$= 4 \sqrt{ \tan^2 \theta . \sin^2 \theta }$

$= 4 \tan \theta \sin \theta$

$\therefore m^2 - n^2 = 4 \sqrt{ mn }$

Hence proved.

$\\$

Question 20: A chord of a circle, of radius $15$ cm, subtends an angle of $60^o$ at the centre of the circle. Find the area of major and minor segments (Take $\pi = 3.14, \sqrt{3} = 1.73$ )

Radius $(r) = 15$ cm $\theta = 60^o$

Therefore Area of $OAPB =$ $\frac{60}{360}$ $\times \pi r^2 =$ $\frac{1}{6}$ $\times 3.14 \times 15^2 = 117.75 \ cm^2$

Area of $\triangle AOB =$ $\frac{1}{2}$ $\times base \times height$

We draw $OM \perp AB$

$\therefore \angle OMB = \angle OMA = 90^o$

In $\triangle OMA$ and $\triangle OMB$

$\angle OMA = \triangle OMB = 90^o$  (by construction)

$OA = OB$   (both are radius of the same circle)

$OM$ is common

$\therefore \triangle OMA \cong \triangle OMB$  (by RHS criterion)

$\Rightarrow \angle AOM = \angle BOM$

$\therefore \angle AOM = \angle BOM =$ $\frac{1}{2}$ $\angle BOA$

$\Rightarrow \angle AOM = \angle BOM =$ $\frac{1}{2}$ $\times 60 = 30^o$

Also, since $\triangle OMB \cong \triangle OMA$

$\therefore BM = AM$

$\Rightarrow BM = AM =$ $\frac{1}{2}$ $AB$

$\Rightarrow AB = 2 BM$  … … … … … i)

In right $\triangle OMB$

$\frac{AM}{AO}$ $= \sin 30^o$

$\Rightarrow$ $\frac{AM}{15}$ $=$ $\frac{1}{2}$

$\Rightarrow AM =$ $\frac{15}{2}$

$\Rightarrow AB = 2 \times$ $\frac{15}{2}$ $= 15$

Similarly, In right $\triangle OMA$

$\frac{OM}{AO}$ $= \cos 30^o$

$\Rightarrow \frac{OM}{15}$ $=$ $\frac{\sqrt{3}}{2}$

$\Rightarrow OM =$ $\frac{\sqrt{3}}{2}$ $\times 15$

$\therefore$ Area of $\triangle AOB =$ $\frac{1}{2}$ $\times 15 \times$ $\frac{\sqrt{3}}{2}$ $\times 15 = 97.3125 \ cm^2$

Therefore area of segment $APB = 117.75 - 97.3125 = 20.4375 \ cm^2$

Area of major segment $= \pi r^2 - 20.4375 = 3.14 \times 15^2 - 20.4375 = 686.0625 \ cm^2$

$\\$

Question 21: A sphere of diameter $12$ cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the vessel rises by $3$ $\frac{5}{9}$ cm. Find the diameter of the cylindrical vessel.

Or

A cylinder whose height is two-third of its diameter, has the same volume as that of a sphere of radius $4$ cm. Find the radius of base of the cylinder.

Volume of sphere $=$ Volume of water displaced

$\frac{4}{3}$ $\pi {r_1}^3 = \pi {r_2}^2 h$

$\Rightarrow$ $\frac{4}{3}$ $\times 6^3 = {r_2}^2 \times$ $\frac{32}{9}$

$\Rightarrow {r_2}^2 =$ $\frac{4}{3}$ $\times$ $\frac{6^3 \times 9}{32}$ $= 81$

$\Rightarrow r_2 = 9$ cm

Therefore diameter  $= 2 \times 9 = 18$ cm

Or

Let the diameter of the cylinder $= x$

Therefore Radius $(r_2) =$ $\frac{x}{2}$ ,   Height $(h) =$ $\frac{2x}{3}$

$\Rightarrow$ $\frac{4}{3}$ $\pi {r_1}^3 = \pi {r_2}^2 h$

$\Rightarrow$ $\frac{4}{3}$ $\times 4^3 = \Big($ $\frac{x}{2}$ $\Big)^2 \times \Big($ $\frac{2x}{2}$ $\Big)$

$\Rightarrow x^3 =$ $\frac{4 \times 4^3 \times 4}{2}$ $= 512$

$\Rightarrow x = 8$ cm

Hence Radius of the base cylinder $(r_2) =$ $\frac{8}{2}$ $= 4$ cm

$\\$

Question 22: The following table gives the daily income of $50$ labourers :

 Daily Income (Rs.): 100-120 120-140 140-160 160-180 180-200 Number of Laborer: 12 14 8 6 10

Find the mean and mode of the above data.

We have

 Daily Income Mid Value $(x_i)$$(x_i)$ Frequency $(f_i)$$(f_i)$ Cumulative Frequency $(cf)$$(cf)$ $f_ix_i$$f_ix_i$ 100-120 110 12 12 1320 120-140 130 14 26 1820 140-160 150 8 34 1200 160-180 170 6 40 1020 180-200 190 10 50 1900 $\Sigma f_i = 50$$\Sigma f_i = 50$ $\Sigma f_ix_i = 7260$$\Sigma f_ix_i = 7260$

Mean $( \overline{x}) =$ $\frac{\Sigma f_ix_i }{\Sigma f_i }$ $=$ $\frac{7260}{50}$ $= 145.2$

$N = 50 \Rightarrow$ $\frac{N}{2}$ $= 25$

Cumulative frequency just greater than $25$ is $26$ and corresponding class is $120-140$

Thus, the Median class is $120-140$

$\therefore L = 120, h = 20, f=14, C = 12,$ $\frac{N}{2}$ $= 25$

Median $(M) = L +$ $\frac{\frac{N}{2} - C}{f}$ $\times h$

$= 120 +$ $\frac{25-12}{14}$ $\times 20$

$= 138.57$

Mode $= 3 M - 2 (\overline{x}) = 3 \times 138.57 - 2 \times 145.2 = 125.31$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: Two taps together can fill a tank in $6$ hours. The tap of larger diameter takes $9$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Or

Solve for $x:$ $\frac{x+1}{x-1}$ $-$ $\frac{x-1}{x+1}$ $=$ $\frac{5}{6}$, $x \neq 1, -1$

Let the time taken by the smaller tap $= x$ hours

Therefore the time taken by the larger tap $= (x-9)$ hours

In 1 hours, smaller tap fills $\frac{1}{x}$ of the tank.

In 1 hours, larger tap fills $\frac{1}{x-9}$ of the tank.

In 1 hours, both taps fills $\frac{1}{6}$ of the tank.

$\therefore$ $\frac{1}{x}$ $-$ $\frac{1}{x-9}$ $=$  $\frac{1}{6}$

$\Rightarrow 6(x - 9 + x) = x(x-9)$

$\Rightarrow 12x - 54 = x^2 - 9x$

$\Rightarrow x^2 - 21x + 54 = 0$

$\Rightarrow x^2 - 18x - 3x +54 = 0$

$\Rightarrow x(x-18) - 3(x-18) = 0$

$\Rightarrow (x-18)(x-3) = 0$

$\Rightarrow x = 18 \ or \ x = 3$

When $x = 18$,  time taken by the larger tap to fill the tank $= 18 -9 = 9$ hours. $x = 3$ is not possible because then the time taken by the larger tap would be negative which is not possible.

Or

$\frac{x+1}{x-1}$ $-$ $\frac{x-1}{x+1}$ $=$ $\frac{5}{6}$

$\Rightarrow$ $\frac{(x+1)^2 - (x-1)^2}{x^2 - 1}$ $=$ $\frac{5}{6}$

$\Rightarrow$ $\frac{x^2 + 1 + 2x - x^2 -1 +2x}{x^2 - 1}$ $=$ $\frac{5}{6}$

$\Rightarrow 4x(6) = 5(x^2 - 1)$

$\Rightarrow 5x^2 - 24x - 5 = 0$

$\Rightarrow (5x+1)(x-5) = 0$

$\Rightarrow x = 5 \ or \ x = -$ $\frac{1}{5}$

$\\$

Question 24: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Or

Prove that in a triangle, if the square of one side is equal to sum of the squares of the other two sides, the angle opposite the first side is a right angle.

Given: $\triangle ABC \sim \triangle DEF$

To Prove: $\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $=$ $\frac{AB^2}{DE^2}$ $=$ $\frac{BC^2}{EF^2}$ $=$ $\frac{AC^2}{DF^2}$

Construction: Draw $AG \perp BC$ and $DH \perp EF$

Proof:

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $=$ $\frac{\frac{1}{2} \times BC \times AG}{\frac{1}{2} \times EF \times DH}$ $=$ $\frac{BC}{EF} \times \frac{AG}{DH}$   … … … … … i)

Now in $\triangle ABG$ and $\triangle DEH$

$\angle B = \angle E$ (given by similarity)

$\angle AGB = \angle DHE = 90^o$ (by construction)

$\therefore \triangle ABG \cong \triangle DEH$ (by AA criterion)

$\therefore \frac{AB}{DE}$ $=$ $\frac{BG}{EH}$ $=$ $\frac{AG}{DH}$

But $\frac{AB}{DE}$ $=$ $\frac{BC}{EF}$  (since $\triangle ABC \sim \triangle DEF$)

$\therefore \frac{AG}{DH}$ $=$ $\frac{BC}{EF}$   … … … … … ii)

From i) and ii)

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $=$ $\frac{BC}{EF} \times \frac{BC}{EF}$ $=$ $\frac{BC^2}{EF^2}$

Similarly, we can prove that

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $=$ $\frac{AB^2}{DE^2}$ $=$ $\frac{AC^2}{DF^2}$

Hence proved

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $=$ $\frac{AB^2}{DE^2} = \frac{BC^2}{EF^2}$ $=$ $\frac{AC^2}{DF^2}$

Or

Given: $AC^2 = AB^2 + BC^2$

To Prove: $\angle B = 90^o$

Construction: $\triangle PQR$ is a right angled at $Q$ such that $PQ = AB$ and $QR = BC$

Proof: From $\triangle PQR$

$PR^2 = PQ^2 + QR^2$ (Pythagoras theorem)

$PR^2 = AB^2 + BC^2$ (by construction)   … … … … … i)

But $AC^2 = AB^2 + BC^2$ (given)   … … … … … ii)

$AC^2 = PR^2$ from i) and ii)

$\therefore AC = PR$   … … … … … iii)

Now in $\triangle ABC$ and $\triangle PQR$

$AB = PQ$ (by construction)

$BC = QR$ (by construction)

$AC = PR$ (from iii)

$\therefore \triangle ABC \cong \triangle PQR$ (by SSS criterion)

$\therefore \angle B = \angle Q$

But $\angle Q = 90^o$ by construction

$\therefore \angle B = 90^o$. Hence proved.

$\\$

Question 25: Write the steps of construction for drawing a $\triangle ABC$ in which
$BC = 8 \ cm, \angle B = 45^o$ and $\angle C = 30^o$. Now write the steps of construction for drawing a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$.

$\\$

Question 26: The sum of the first n terms of an A.P. is $5n2 + 3n$. If its $m^{th}$ term is $168$, find the value of $m$. Also find the $20^{th}$ term of the A.P.

Or

The $4^{th}$ and the last terms of an A.P. are $11$ and $89$ respectively. If there are $30$ terms in the A.P., find the A.P. and its $23^{rd}$ term.

Sum of the first n terms $= S_n$

$\therefore S_n = 5n^2 + 3n$

For $n=1, S_1 = 5(1)^2 + 3(1) = 8 = t_1$

For $n_2, S_2 = 5(2)^2 + 3(2) = 26$

$\therefore t_2 = S_2 - S_1 = 26-8 = 18$

$\therefore$ common difference $= t_2 - t_1 = 18-8 = 10$

$\therefore a = 8, d = 10$

Therefore $t_m = a + (m-1) d$

$\Rightarrow 168 = 8 + (m-1) (10)$

$\Rightarrow 160 = (m-1) (10)$

$\Rightarrow 16 = m - 1$

$\Rightarrow m = 17$

$t_{20} = a + (20-1) d$

$\Rightarrow t_{20} = 8 + (19)(10)$

$\Rightarrow t_{20} = 198$

Hence $m = 17$ and $t_{20} = 198$

Or

The $n^{th}$ of an AP where the first term is $a$ and the common difference is $d$

$a_n = a + (n-1)d$

$\Rightarrow a_4 = a + (4-1)d$

$\Rightarrow 11 = a + 3d$  … … … … … i)

Also $a_{30} = a + (30-1) d$

$\Rightarrow 89 = a + 29 d$   … … … … … ii)

Subtracting i) from ii) we get

$(89-11) = (29-3)d \Rightarrow d =$ $\frac{78}{16}$ $= 3$

$\therefore a = 11 - 3(3) = 2$

Therefore AP is $2, 5, 8, 11, \ldots$

$a_{23} = a + (23-1)d$

$\Rightarrow a_{23} = 2 + 22 \times 3 = 68$

Hence the AP is $2, 5, 8, 11, \ldots$ and $a_{23} = 68$

$\\$

Question 27: Prove that: $\Big($ $\frac{\sin A}{1 - \cos A}$ $-$ $\frac{1 - \cos A}{\sin A}$ $\Big) . \Big($ $\frac{\cos A}{1 - \sin A}$ $-$ $\frac{1 - \sin A}{\cos A}$ $\Big) = 4$

LHS $= \Big($ $\frac{\sin A}{1 - \cos A}$ $-$ $\frac{1 - \cos A}{\sin A}$ $\Big) . \Big($ $\frac{\cos A}{1 - \sin A}$ $-$ $\frac{1 - \sin A}{\cos A}$ $\Big)$

$= \Big[$ $\frac{\sin^2 A - (1+ \cos^2 A - 2 \cos A)}{(1- \cos A) \sin A}$ $\Big] \Big[$ $\frac{\cos^2 A - (1 + \sin^2 A - 2 \sin A)}{(1- \sin A) \cos A}$ $\Big]$

$= \Big[$ $\frac{\sin^2 A - 1- \cos^2 A + 2 \cos A)}{(1- \cos A) \sin A}$ $\Big] \Big[$ $\frac{\cos^2 A - 1 - \sin^2 A + 2 \sin A}{(1- \sin A) \cos A}$ $\Big]$

$= \Big[$ $\frac{ 2 \cos A - 2 \cos^2 A}{(1- \cos A) \sin A}$ $\Big] \Big[$ $\frac{2 \sin A - 2 \sin^2 A}{(1- \sin A) \cos A}$ $\Big]$

$= \Big[$ $\frac{ 2 \cos A (1 - \cos A)}{(1- \cos A) \sin A}$ $\Big] \Big[$ $\frac{2 \sin A( 1 - \sin A)}{(1- \sin A) \cos A}$ $\Big]$

$= 2 \times 2$

$= 4 =$ RHS

Hence Proved.

$\\$

Question 28: A statue, $1.46$ m tall, stands on a pedestal. From a point on the ground the angle of elevation of the top of the statue is $60^o$ and from the same point angle of elevation of the top of the pedestal is $45^o$. Find the height of the pedestal. (use $\sqrt{3} = 1.73$)

In $\triangle ABC$

$\frac{BC}{AB}$ $= \tan 45^o = 1 \Rightarrow AB = BC$   … … … … … i)

From $\triangle ABD$

$\frac{DB}{AB}$ $= \tan 60^o = \sqrt{3}$

$\Rightarrow \frac{1.46 + BC}{AB}$ $= \tan 60^o = \sqrt{3}$

$\Rightarrow 1.46 + BC = \sqrt{3} AB$   … … … … … ii)

Substituting i) and ii)  we get

$1.46 + BC = \sqrt{3} BC$

$\Rightarrow BC =$ $\frac{1.46}{\sqrt{3} - 1}$ $=$ $\frac{1.46}{1.73 - 1}$ $= 2$ m

Therefore the height of the pedestal $= 2$ m

$\\$

Question 29: Sudhakar donated $3$ cylindrical drums to store cereals to an orphanage. If radius of each drum is $0.7$ m and height $2$ m, find the volume of each drum. If each drum costs Rs. $350 \ per \ m^3$, find the amount spent by Sudhakar for orphanage. What value is exhibited in the question. (Use $\pi =$ $\frac{22}{7}$)

Radius $(r)$ of drum $= 0.7$ m

Height $(h)$ of the drum $= 2$ m

Volume of each drum $= \pi r^2 h =$ $\frac{22}{7}$ $\times {0.7}^2 \times 2 = 3.08 \ m^3$

Cost of drum $= 350$ $\frac{Rs}{m^3}$

Therefore amount spent $= 3 \times 3.08 \times 350 = 3234$ Rs.

$\\$

Question 30: The median of the following data is $52.5$. If the total frequency is $100$, find the values of $x$ and $y$.

 Classes Frequency 0-10 2 10-20 5 20-30 $x$$x$ 30-40 12 40-50 17 50-60 20 60-70 7 70-80 9 80-90 7 90-100 4

 Classes Frequency $(f_i)$$(f_i)$ Cumulative Frequency $(cf)$$(cf)$ 0-10 2 2 10-20 5 7 20-30 $x$$x$ 7 $+x$$+x$ 30-40 12 19 $+x$$+x$ 40-50 17 36 $+x$$+x$ 50-60 20 56 $+x$$+x$ 60-70 $y$$y$ 56 $+x+y$$+x+y$ 70-80 9 65 $+x+y$$+x+y$ 80-90 7 72 $+x+y$$+x+y$ 90-100 4 76 $+x+y$$+x+y$

$76+x+y = 100 \Rightarrow x + y = 24$    … … … … … i)

Median class $= 50 -60$

$L = 50,$ $\frac{N}{2}$ $= 25, cf = 36+x, C = 10, f = 20, Median = 52.5$

Median $(M) = L +$ $\frac{\frac{N}{2} - cf}{f}$ $\times C$

$\Rightarrow 52.5 = 50 + \Big($ $\frac{50-36-x}{20}$ $\Big) \times 10$

$\Rightarrow 52.5 = 50 +$ $\frac{14-x}{2}$

$\Rightarrow 52.5 = 50 + 7 -$ $\frac{x}{2}$

$\Rightarrow$ $\frac{x}{2}$ $= 57 - 52.5 = 4.5$

$\Rightarrow x = 2 \times 4.5 = 9$

$\therefore y = 24 - 9 = 15$

Hence $x = 9$ and $y = 15$

$\\$


text:
# Write down the assumptions of interpolation and extrapolation.

278 views

Write down the assumptions of interpolation and extrapolation.

+1 vote
by (64.3k points)
selected

• The assumptions made in interpolation and extrapolations are:
• There are no sudden jumps in the values of dependent variable(Y) from one period to another(X).
• The rate of change of figures (Y) from one period to another(X) is uniform.
• There will be no consecutive missing values in the series.

format: plain


text:
# Function from Mathlab

#1

Good morning,
I really do not know where to put this question.
Is there a package that contains a function similar to Mathlab's chebfun2v function?
I need your guidance, both in l function and where to make this type of question that has nothing to do with RStudio.
regards
Hermes
Gracias

#2

The #general category (where you put this post) is a great choice!

I don't know the answer to this part. It might help if you explain what you specifically are using chebfun2v objects for in MATLAB, since there may not be an identical implementation in R, but somebody may know of tools that accomplish the same goal.

You might also take a look at the CRAN Task View for Numerical Mathematics, which is a guide to the R packages in this area: https://cran.r-project.org/web/views/NumericalMathematics.html

For others who may know more R than MATLAB, Chebfun is a MATLAB package for numerical computing:

Chebfun is an open-source package for computing with functions to about 15-digit accuracy. Most Chebfun commands are overloads of familiar MATLAB commands — for example sum(f) computes an integral, roots(f) finds zeros, and u = L\f solves a differential equation.

And chebfun2v specifically:

Chebfun2 can represent scalar-valued functions, such as \exp(x+y), and vector-valued functions, such as [\exp(x+y);\cos(x−y)]. A vector-valued function is called a chebfun2v, and chebfun2v objects are useful for computations of vector calculus.

#3

Thank you for your guidance and clarify me about the object chebfun2v (objects are useful for computations of vector calculus).
my interest is to reproduce the examples of the following site:
http://www.chebfun.org/examples/ode-linear/DynamicalSystems.html
And although I have been studying and reproducing examples, found on the web, of the Solve and PhaseR package for months, I still do not feel very familiar with them.
The function or object chebfun2v seems to me a "black box". Thanks to his explanation I am orienting myself.


text:
Home > Research > Publications & Outputs > Measurement of the ratio of cross sections for ...

## Measurement of the ratio of cross sections for inclusive isolated-photon production in pp collisions at √s = 13 and 8 TeV with the ATLAS detector

Research output: Contribution to Journal/MagazineJournal articlepeer-review

Published

### Standard

In: Journal of High Energy Physics, Vol. 2019, No. 4, 93, 01.04.2019.

Research output: Contribution to Journal/MagazineJournal articlepeer-review

### Bibtex

@article{e42eb6c55f0d48be92dde9273376d0e9,
title = "Measurement of the ratio of cross sections for inclusive isolated-photon production in pp collisions at √s = 13 and 8 TeV with the ATLAS detector",
abstract = "The ratio of the cross sections for inclusive isolated-photon production in pp collisions at centre-of-mass energies of 13 and 8 TeV is measured using the ATLAS detector at the LHC. The integrated luminosities of the 13 TeV and 8 TeV datasets are 3.2 fb −1 and 20.2 fb −1 , respectively. The ratio is measured as a function of the photon transverse energy in different regions of the photon pseudorapidity. The predictions from next-to-leading-order perturbative QCD calculations are compared with the measured ratio. The experimental systematic uncertainties as well as the uncertainties affecting the predictions are evaluated taking into account the correlations between the two centre-of-mass energies, resulting in a reduction of up to a factor of 2.5 (5) in the experimental (theoretical) systematic uncertainties. The predictions based on several parameterisations of the proton parton distribution functions agree with the data within the reduced experimental and theoretical uncertainties. In addition, this ratio to that of the fiducial cross sections for Z boson production at 13 and 8 TeV using the decay channels Z → e + e − and Z → μ + μ − is made and compared with the theoretical predictions. In this double ratio, a further reduction of the experimental uncertainty is obtained because the uncertainties arising from the luminosity measurement cancel out. The predictions describe the measurements of the double ratio within the theoretical and experimental uncertainties.[Figure not available: see fulltext.]. {\textcopyright} 2019, The Author(s).",
author = "ATLAS Collaboration and A.E. Barton and I.A. Bertram and G. Borissov and E.V. Bouhova-Thacker and H. Fox and R.C.W. Henderson and R.W.L. Jones and V. Kartvelishvili and R.E. Long and P.A. Love and D. Muenstermann and A.J. Parker and M. Smizanska and A.S. Tee and J. Walder and A.M. Wharton and B.W. Whitmore",
year = "2019",
month = apr,
day = "1",
doi = "10.1007/JHEP04(2019)093",
language = "English",
volume = "2019",
journal = "Journal of High Energy Physics",
issn = "1029-8479",
publisher = "Springer-Verlag",
number = "4",

}

### RIS

TY - JOUR

T1 - Measurement of the ratio of cross sections for inclusive isolated-photon production in pp collisions at √s = 13 and 8 TeV with the ATLAS detector

AU - Collaboration, ATLAS

AU - Barton, A.E.

AU - Bertram, I.A.

AU - Borissov, G.

AU - Bouhova-Thacker, E.V.

AU - Fox, H.

AU - Henderson, R.C.W.

AU - Jones, R.W.L.

AU - Kartvelishvili, V.

AU - Long, R.E.

AU - Love, P.A.

AU - Muenstermann, D.

AU - Parker, A.J.

AU - Smizanska, M.

AU - Tee, A.S.

AU - Walder, J.

AU - Wharton, A.M.

AU - Whitmore, B.W.

PY - 2019/4/1

Y1 - 2019/4/1

N2 - The ratio of the cross sections for inclusive isolated-photon production in pp collisions at centre-of-mass energies of 13 and 8 TeV is measured using the ATLAS detector at the LHC. The integrated luminosities of the 13 TeV and 8 TeV datasets are 3.2 fb −1 and 20.2 fb −1 , respectively. The ratio is measured as a function of the photon transverse energy in different regions of the photon pseudorapidity. The predictions from next-to-leading-order perturbative QCD calculations are compared with the measured ratio. The experimental systematic uncertainties as well as the uncertainties affecting the predictions are evaluated taking into account the correlations between the two centre-of-mass energies, resulting in a reduction of up to a factor of 2.5 (5) in the experimental (theoretical) systematic uncertainties. The predictions based on several parameterisations of the proton parton distribution functions agree with the data within the reduced experimental and theoretical uncertainties. In addition, this ratio to that of the fiducial cross sections for Z boson production at 13 and 8 TeV using the decay channels Z → e + e − and Z → μ + μ − is made and compared with the theoretical predictions. In this double ratio, a further reduction of the experimental uncertainty is obtained because the uncertainties arising from the luminosity measurement cancel out. The predictions describe the measurements of the double ratio within the theoretical and experimental uncertainties.[Figure not available: see fulltext.]. © 2019, The Author(s).

AB - The ratio of the cross sections for inclusive isolated-photon production in pp collisions at centre-of-mass energies of 13 and 8 TeV is measured using the ATLAS detector at the LHC. The integrated luminosities of the 13 TeV and 8 TeV datasets are 3.2 fb −1 and 20.2 fb −1 , respectively. The ratio is measured as a function of the photon transverse energy in different regions of the photon pseudorapidity. The predictions from next-to-leading-order perturbative QCD calculations are compared with the measured ratio. The experimental systematic uncertainties as well as the uncertainties affecting the predictions are evaluated taking into account the correlations between the two centre-of-mass energies, resulting in a reduction of up to a factor of 2.5 (5) in the experimental (theoretical) systematic uncertainties. The predictions based on several parameterisations of the proton parton distribution functions agree with the data within the reduced experimental and theoretical uncertainties. In addition, this ratio to that of the fiducial cross sections for Z boson production at 13 and 8 TeV using the decay channels Z → e + e − and Z → μ + μ − is made and compared with the theoretical predictions. In this double ratio, a further reduction of the experimental uncertainty is obtained because the uncertainties arising from the luminosity measurement cancel out. The predictions describe the measurements of the double ratio within the theoretical and experimental uncertainties.[Figure not available: see fulltext.]. © 2019, The Author(s).

KW - Photon production

KW - QCD

U2 - 10.1007/JHEP04(2019)093

DO - 10.1007/JHEP04(2019)093

M3 - Journal article

VL - 2019

JO - Journal of High Energy Physics

JF - Journal of High Energy Physics

SN - 1029-8479

IS - 4

M1 - 93

ER -


text:
# Relative entropy and squashed entanglement

Research output: Contribution to journalArticleResearchpeer-review

23 Citations (Scopus)

## Abstract

© Springer-Verlag Berlin Heidelberg 2014. We are interested in the properties and relations of entanglement measures. Especially,we focus on the squashed entanglement and relative entropy of entanglement, as well as their analogues and variants. Our first result is a monogamy-like inequality involving the relative entropy of entanglement and its one-way LOCC variant. The proof is accomplished by exploring the properties of relative entropy in the context of hypothesis testing via one-way LOCC operations, and by making use of an argument resembling that by Piani on the faithfulness of regularized relative entropy of entanglement. Following this, we obtain a commensurate and faithful lower bound for squashed entanglement, in the form of one-way LOCC relative entropy of entanglement. This gives a strengthening to the strong subadditivity of von Neumann entropy. Our result improves the trace-distance-type bound derived in Brandão et al. (Commun Math Phys, 306:805–830, 2011), where faithfulness of squashed entanglement was first proved. Applying Pinsker’s inequality, we are able to recover the trace-distance-type bound, even with slightly better constant factor. However, the main improvement is that our new lower bound can be much larger than the old one and it is almost a genuine entanglement measure. We evaluate exactly the relative entropy of entanglement under various restricted measurement classes, for maximally entangled states. Then, by proving asymptotic continuity, we extend the exact evaluation to their regularized versions for all pure states. Finally,we consider comparisons and separations between some important entanglement measures and obtain several new results on these, too.
Original language English 63-80 Communications in Mathematical Physics 326 https://doi.org/10.1007/s00220-013-1871-2 Published - 25 Jan 2014


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# Silicone Rubber part, behavior and wall thickness3

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## Silicone Rubber part, behavior and wall thickness

(OP)
Dear forum members,

I am working on a silicone part for a new product. You should be able to push it down, and then it should stay there, until you pull it back up. I have no clue how to choose the exact shape and wall thickness, so that it is easy to do this with one hand, maybe I should have a variable wall thickness or other features for the desired functionality. I chose silicone rubber as a material, because of the fact that is will be incorporated in a "kitchen appliance" and it should withstand the dishwasher.

My explanation might not be clear enough so I have included images.

Maybe there is some way to simulate this phenomenon, but I reckon it's too complicated and I'm better off with an expert's opinion from someone who works with rubber daily.

Thanks you,

Max

(OP)

(OP)

### RE: Silicone Rubber part, behavior and wall thickness

(OP)
Is there no one who can help me with this, or did I use the wrong forum?

Max

### RE: Silicone Rubber part, behavior and wall thickness

I think you've posted it in the right forum Max but there's no-one able to help. I suggest you start by talking to your silicone supplier's technical department.

### RE: Silicone Rubber part, behavior and wall thickness

Yeah, don't know of any forum that would work any better. The phenomena you describe is called "snap through" buckling, and it is a pretty non-linear problem to work on. Personally, I don't think your proposed shape will work, I think the outer edge has to roll over for the deformed condition to be stable (unless the bottom gets "stuck" to something?). The "poppers" that my kids played with looked similar to what you drew, but they were not stable in the buckled position, and would pop back after several seconds, hopping into the air when they did so.

The good news is, silicones are easy to mold. If you have access to a lathe, you should be able to bang out prototype mold halves (top and bottom) and tweak the wall thickness and outer edge detail (you may want a thick "bead" at the o.d. to control shape, but again, I'm just guessing). Silicone mold compounds are readily available in small quantities. Check Mcmaster-carr.

### RE: Silicone Rubber part, behavior and wall thickness

(OP)
Thanks for the replies. I had contacted a silicon supplier, but they're not very helpful, so far. I'm going to try and make some mock-up models with a sugru substitute (silicone chaulk and corn starch). And later maybe some higher quality parts with casting silicon. Thanks for the description of snap through buckling, it seems as though I might need to add some "fold lines", or look at collapsible silicone tableware.

Max

### RE: Silicone Rubber part, behavior and wall thickness

You're welcome for what little help we can give, Max. I think your fastest route forward is to play with real objects, so you are going the right way. Do be careful, silicones are available in a wide variety of stiffnesses/durometers and you can change the way your part behaves by changing compounds. But, with enough experimentation, you may come up with some design rules of your own on how the shape needs to change to compensate for material variations...

Good luck.

### RE: Silicone Rubber part, behavior and wall thickness

Max,

My suggestion would be to use Finite Element Analysis. You will need a non-linear package such as Abaqus and some tensile data on a material.

An experienced analyst would be able to evaluated multiple design concepts in a short amount of time.

My experience is that analysis is faster, cheaper, and provides more detail than going straight to prototypes.

Good Luck

### RE: Silicone Rubber part, behavior and wall thickness

"and some tensile data on a material."

I like FEA, but predicting non-linear snap-through buckling on non-linear material? You're talking pretty cutting edge stuff there, people are writing their PhD theses on these types of topics.

You need a lot more that just a simple pull test to even approximately characterise rubber. And every rubber formulation is different, so you need test coupons for each one. So, he would have to buy samples of rubber and cast coupons, and send them to a lab, and do a series of FEA runs on several geometries. What's a seat of Abaqus or ANSYS cost, with full non-linear solver packages, upwards of $20k per seat? "My experience is that analysis is faster, cheaper, and provides more detail than going straight to prototypes." I could build a prototype mold for less than$1000...I'm not sure the analysis approach is the best way for this project.

### RE: Silicone Rubber part, behavior and wall thickness

btrueblood,

Your right there is a lot to it and maybe the cost is too high. I was simply throwing out an option that I have used as a product engineer (at a cost sensitive manufacturer) and have expertise in as an analyst.

1. The goal, as I see it, isn't a deadnuts match between analysis and product but a series of simple analyses (2d, axisymmetric) to point in a direction. Then onto to prototyping.
2. I have modeled parts that would go through a wall collapse or snap behavior. If a static run didn't work then onto quasi-static dynamic runs.
3. Abaqus has a material model called Marlow that takes one tensile data set to characterize the material. The Marlow model does as well as or better than the three expensive and difficult tests recommended by literature and used by other material models.
4. Yes, each elastomer is different. The goal is to get to a part behavior and you could do that with one material. Prototype to confirm and maybe learn enough so that if you have to switch materials then another material test and analysis may not be needed.
5. If the op does not have analysis resources already in house then he would have to contract it out to make the cost palatable. You wouldn't buy a seat nor spend the time needed to learn it for one project.
6. At my previous job, the prototype molds cost well beyond a \$1000 a piece. If you could prevent building a couple prototype molds along with the time savings of waiting for the mold to be built, waiting for press time and then figuring out what to do next when things didn't work out, and repeat, well, analysis ending up saving time and money.

### RE: Silicone Rubber part, behavior and wall thickness

(OP)
Hi all,

Thanks for the in-depth discussion.

I've messed around with some silicon and made a quick and dirty prototype that luckily suggests that the desired behaviour is quite easy to obtain. I'm going ahead and order a prototype for my total product. I can make some final adjustments afterwards if needed. It helps that the product is not extremely cost-sensitive. Maybe I'll even ask my Abaqus trained friends to help me out with some additional simulation.

Cheers, Max

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text:
Show that
Question:

A. $\frac{\pi}{3}$

B. $\frac{2 \pi}{3}$

C. $\frac{\pi}{6}$

D. $\frac{\pi}{12}$

Solution:

$\int \frac{d x}{1+x^{2}}=\tan ^{-1} x=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

$=\tan ^{-1} \sqrt{3}-\tan ^{-1} 1$

$=\frac{\pi}{3}-\frac{\pi}{4}$

$=\frac{\pi}{12}$

Hence, the correct answer is D.


text:
# No need for gravity assist with Falcon Heavy?

Elon Musk is quoted to have said, after the launch of Falcon Heavy, that «it can launch things direct to Pluto or beyond, no stop needed, don’t even need gravity assist or anything». Now, I am not an expert in orbital computation, but this claim seems highly dubious. New horizon took 9 years to reach Pluto. How long would it take for a spacecraft launch by Falcon Heavy without using any gravity assist?

• Long time and/or small probe. That it would be possible does not mean it would be a good idea. – jkavalik Feb 7 '18 at 20:16
• New Horizons was launched by an Atlas V 551, which has a payload to GEO of 8.9 tons. FH has 26.7 tons of payload to GEO. That difference translates into a lot of extra delta-V for a small payload. – Hobbes Feb 7 '18 at 20:22
• @Hobbes Delta v is logarithmic in the ratio of initial to final mass, no? Your figures would result in a log 3 at best, hardly 10% Am I missing something in your reasoning? – frapadingue Feb 7 '18 at 20:31
• Too tired to do the calculations now. – Hobbes Feb 7 '18 at 20:42
• 26.7t to GTO, not GEO. – Russell Borogove Feb 7 '18 at 20:49

It could have done so, but it is not the only rocket that can do so. The Atlas V that launched New Horizons to Pluto could have taken a direct path, but it would have taken 3 years longer. Falcon Heavy reportedly can lift 3.5 tonnes to Pluto directly, which is about 7 times heavier then New Horizons.

As to how quickly could Falcon Heavy have done that without the gravity assist? The exact numbers are pretty difficult to guess. SpaceX no longer publishes their performance charts like they used to. From NASA's performance estimator, I have the following charts. Note that the performance numbers are suspect, will update them when I have better.

This indicates that while a direct to Pluto mission is possible, it would not have the high energy required. Even with the expendable version, it still doesn't perform as well.

Bottom line is, the lack of a high ISP upper stage really limits the functionality of the Falcon Heavy in carrying cargo beyond Earth orbit. The mission could be done, but it would take longer then with an Atlas V 551 mission, as New Horizons used. For a fully expendable mission, the crossover point between Falcon Heavy and Atlas V 551 seems to be about $60 km^2/s^2$. That won't quite get you to Jupiter. It turns out the two are pretty equivalent in price.

• OTOH, couldn't Falcon Heavy take a fairly big probe with a lot of xenon for a ion engine to LEO instead? – SF. Feb 12 '18 at 14:12
• Sure, or some other similar other high ISP type system. That doesn't usually work very well for getting to the outer solar system, however. – PearsonArtPhoto Feb 12 '18 at 14:22
• It seems those graphs were generated by the KSC Performance Calculator: elvperf.ksc.nasa.gov/Pages/Query.aspx According to an NSF post, that calculator uses outdated information (forum.nasaspaceflight.com/…) – Hobbes Feb 12 '18 at 14:38
• Well, I can't find anything better... Will give it another shot when I get a chance. Sigh... – PearsonArtPhoto Feb 12 '18 at 14:55
• @ORcoder Good point, I've updated accordingly. – PearsonArtPhoto Feb 16 '18 at 22:46


text:
### Session D38: Nanoinstrumentation for Biological and Other Applications

2:30 PM–5:06 PM, Monday, March 5, 2007

Chair: Karen Waldrip, Sandia National Laboratories

Abstract ID: BAPS.2007.MAR.D38.3

### Abstract: D38.00003 : Dynamic Detection of a Single Bacterium: Nonlinear Rotation Rate Shifts of Driven Magnetic Microspheres

2:54 PM–3:06 PM

MathJax On | Off     Abstract

#### Authors:

Brandon H. McNaughton
(University of Michigan)

Rodney R. Agayan
(University of Michigan)

Raoul Kopelman
(University of Michigan)

We report on a new technique which was used to detect single Escherichia coli that is based on the changes in the nonlinear rotation of a magnetic microsphere driven by a magnetic field. The presence of one Escherichia Coli bacterium on the surface of a 2.0 micron magnetic microsphere (with an aluminum nanocap'' that indicates the microsphere's orientation) caused an easily measurable change in the drag of the system and, therefore, in the nonlinear rotation rate. The straight-forward measurement uses standard microscopy techniques and the observed average shift in the nonlinear rotation frequency changed by a factor of $\sim$3.8 (Arxiv preprint cond-mat/0610144). Further miniaturization will allow for dynamic detection of viruses and potentially even biomolecules in fluidic environments.

To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2007.MAR.D38.3


text:
# Solve of cubic and quartic equations too slow

In my problem I have a third order algebraic equation for the variable sigma, all other letters are parameters. Here is it's right-hand side, left hand side is zero (you can copy it into your programm):

eq = Er k^2 γ ξ (1 + λ Cos[2 ϕ]) (3 I k Cos[ϕ] + 4 σ Cos[2 ϕ] +
I k Cos[3 ϕ]) Subscript[α, ac] + (ξ + σ) (σ + I k Cos[ϕ]) (8 k^4 +
16 k^2 tf1 + 2 k^4 γ + k^4 γ λ^2 + 8 Er k^2 γ σ + 16 Er tf1 γ σ +
4 k^4 γ λ Cos[2 ϕ] + k^4 γ λ^2 Cos[4 ϕ] +
4 k^2 (k^2 + Er γ σ) Sin[2 ϕ]^2 Subscript[μ, 1] +
4 (k^2 + Er γ σ) (k^2 + tf1 - tf1 Cos[2 ϕ]) Subscript[μ, 2])

My purpose here is to get Taylor expansion in parameter k for all the roots near k=0:

sol = Solve[eq == 0, σ]
sol = σ /. sol
Series[sol[[1]], {k, 0, 2}

The thing is, that evalution of the line sol = Solve[eq == 0, σ] takes really long time. It takes couple minutes here, but when I am trying to solve similar equation of the 4th order, it takes forever. Note, that in the line sol = Solve[eq == 0, σ], I didn't ask to simplify. But looks like mathematica silently simplifies it, since it is running for a long time. Note, that for the third and forth order equations exact algebraic expressions for the solutions are availible, and indeed, if I type something like Solve[a*x^3+b*x^2+c*x+d==0,x], it gives me the answer instantly. It should be really fast procedure to substitute corresponding expressions into the final formula.

Is there a way to get Taylor expansion I need faster? Solving the equations takes much longer time, than Taylor series.

In case someone needs, this is the 4th order equation, which I cannot solve even in 30 minutes:

-Er k^2 γ ξ (1 + λ Cos[2 ϕ]) (4 (k^2 + 2 σ (σ + τ1)) Cos[2 ϕ] + k^2 (3 + Cos[4 ϕ]))
Subscript[α, ac] - (ξ + σ) (k^2 + 2 σ (σ + τ1) + k^2 Cos[2 ϕ]) (8 k^4 + 16 k^2 tf1 +
2 k^4 γ + k^4 γ λ^2 + 8 Er k^2 γ σ + 16 Er tf1 γ σ + 4 k^4 γ λ Cos[2 ϕ] +
k^4 γ λ^2 Cos[4 ϕ] + 4 k^2 (k^2 + Er γ σ) Sin[2 ϕ]^2 Subscript[μ, 1] +
4 (k^2 + Er γ σ) (k^2 + tf1 - tf1 Cos[2 ϕ]) Subscript[μ, 2])

Evaluation of sol = Solve[eq == 0,σ] takes forever.

I found out, that Defer can be used to supress evalution, but I need a different thing.

Thanks, MIkhail

• I find that sol = Solve[eq == 0, σ] takes just seconds, although the result is rather large. By the way, avoid using Subscript. – bbgodfrey Dec 9 '15 at 0:20
• Furthermore, solving the fourth order equation takes only a minute or so. How much memory do you have on your computer? – bbgodfrey Dec 9 '15 at 0:32
• sol = Solve[eq == 0, σ] took 2 minutes on my computer. The method I presented below takes around 3 seconds. – JungHwan Min Dec 9 '15 at 1:01
• bbgodfrey, what PC do you have? I have laptop with 8GB RAM, quadcore intel i7. Using device manager I found out, that during the calculations ~20% of CPU and ~40% of RAM is used, so I doubt, that the issue is in memory. Thanks for the edit, btw – Mikhail Genkin Dec 9 '15 at 1:26
• @MikhailGenkin I have an essentially identical computer running 10.3.0 for Microsoft Windows (64-bit) (October 9, 2015). The calculations complete so quickly that I cannot accurately determine CPU and memory usage with Task Manager for the cubic case. The quartic case takes less than a minute with 17% CPU usage and about 200 MB memory usage. – bbgodfrey Dec 9 '15 at 1:36

Here is a workaround:

eq = Er k^2 γ ξ (1 + λ Cos[2 ϕ]) (3 I k Cos[ϕ] + 4 σ Cos[2 ϕ] +
I k Cos[3 ϕ]) Subscript[α, ac] + (ξ + σ) (σ + I k Cos[ϕ]) (8 k^4 +
16 k^2 tf1 + 2 k^4 γ + k^4 γ λ^2 + 8 Er k^2 γ σ + 16 Er tf1 γ σ +
4 k^4 γ λ Cos[2 ϕ] + k^4 γ λ^2 Cos[4 ϕ] +
4 k^2 (k^2 + Er γ σ) Sin[2 ϕ]^2 Subscript[μ, 1] +
4 (k^2 + Er γ σ) (k^2 + tf1 - tf1 Cos[2 ϕ]) Subscript[μ, 2]);
coeffs = Reverse@CoefficientList[eq, σ];
sol = σ /. Solve[FromDigits[Array[#[] &, Length[coeffs], 1],
σ] == 0, σ] /. n_Integer[] :> coeffs[[n]];
Series[sol[[1]], {k, 0, 2}]

Instead of putting eq in Solve, I put a polynomial with dummy variables (1[], 2[], etc.) and then replaced the dummies with the corresponding coefficients in eq.

It seems that this method takes about 3 seconds on my computer when I first ran it; 0.03 seconds for subsequent iterations.

The bottom 3 lines (from coeffs = to the end of code) should work for any equation eq.

• Thanks a lot, you method really works fast on my laptop! Can you briefly explain, why? But this thing definitely works for me. I will accept your answer it 2 days. – Mikhail Genkin Dec 9 '15 at 1:23
• Nevermind, I got it. The reason I didn't figure it out by myself, is because I thought, that Solve[eq==0,s] works like this. It seems that @bbgodfrey has a new version, which is doing it by default. (He wrote, that his mathematica is solving it very fast). I assume, that you too have an older version of Mathematica, like I do. My is 10.0.1.0 – Mikhail Genkin Dec 9 '15 at 1:43
• Nice but might be overly complicated. To see what I mean, compare constant terms with what you get from Solve[(eq /. k -> 0) == 0, \[Sigma]]. Much smaller and in agreement with the larger result after some random substitution. – Daniel Lichtblau Dec 9 '15 at 15:12

I would go about this differently. You have, in effect, a curve parametrized by sigma and k (regard the other parameters as fixed for the moment). This curve is the zero set for eq. You are interested in expanding sigma as a series in k, around k=0. This can be done by setting up equations for derivatives of sigma regarded explicitly as a function of k; these equations arise from differentiating the original one. As we are evaluating derivatives at k=0 we can make that substitution after differentiating but before solving. Here is the process.

eq = Er k^2 γ ξ (1 + λ Cos[
2 ϕ]) (3 I k Cos[ϕ] + 4 σ Cos[2 ϕ] +
I k Cos[3 ϕ]) Subscript[α,
ac] + (ξ + σ) (σ + I k Cos[ϕ]) (8 k^4 +
16 k^2 tf1 + 2 k^4 γ + k^4 γ λ^2 +
8 Er k^2 γ σ + 16 Er tf1 γ σ +
4 k^4 γ λ Cos[2 ϕ] +
k^4 γ λ^2 Cos[4 ϕ] +
4 k^2 (k^2 + Er γ σ) Sin[
2 ϕ]^2 Subscript[μ, 1] +
4 (k^2 + Er γ σ) (k^2 + tf1 -
tf1 Cos[2 ϕ]) Subscript[μ, 2]);
eqk = eq /. σ -> σ[k];
derivsys = Table[D[eqk, {k, j}], {j, 0, 3}];
derivvars = Table[D[σ[k], {k, j}], {j, 0, 3}];
newvars = Array[d, 4, 0];
sys = (derivsys /. subs) /. k -> 0;
soln = Solve[sys == 0, newvars];

During evaluation of In[425]:= Solve::svars: Equations may not give solutions for all "solve" variables. >>

In[436]:= Most[newvars].(k^Range[0, 2]/Factorial[Range[0, 2]]) /. soln

(* Out[436]= {(k^2 Sec[ϕ] (16 tf1 Cos[ϕ] +
3 Er γ Cos[ϕ] Subscript[α, ac] +
3 Er γ λ Cos[ϕ] Cos[
2 ϕ] Subscript[α, ac] +
Er γ Cos[3 ϕ] Subscript[α, ac] +
Er γ λ Cos[2 ϕ] Cos[
3 ϕ] Subscript[α, ac] +
4 tf1 Cos[ϕ] Subscript[μ, 2] -
4 tf1 Cos[ϕ] Cos[2 ϕ] Subscript[μ,
2]))/(4 Er tf1 γ (-4 - Subscript[μ, 2] +
Cos[2 ϕ] Subscript[μ, 2])), -I k Cos[ϕ] + (
k^2 (1 + λ Cos[2 ϕ]) (-3 Cos[ϕ] +
4 Cos[ϕ] Cos[2 ϕ] -
Cos[3 ϕ]) Sec[ϕ] Subscript[α, ac])/(
4 tf1 (-4 - Subscript[μ, 2] +
Cos[2 ϕ] Subscript[μ, 2])), -ξ - (
k^2 Cos[2 ϕ] (1 + λ Cos[2 ϕ]) Subscript[α,
ac])/(tf1 (-4 - Subscript[μ, 2] +
Cos[2 ϕ] Subscript[μ, 2]))} *)

Astute readers will notice there was a wrinkle. Due to multiplicity and/or factors of sigma in the reduced derivative equations (by reduced I mean with the k->0 substitution), we get underdetermined systems. A way around this is to take more derivatives than we actually need, enough to get solutions for the lower derivatives that we require in the expansion. Above I used one extra derivative. I'm sure there is a theory behind this need for extra derivatives (some aspect of prolongation, maybe) but I do not know the details at all.

• Thanks for the interesting method. Your code seems to be incorrect: my result is different. You never use variable derivs – Mikhail Genkin Dec 9 '15 at 19:04
• (1) Thanks for spotting that mistake. I had some earlier definitions and decided to change notation; what got posted was a bit of a mixture. I believe I have corrected it now. – Daniel Lichtblau Dec 9 '15 at 19:41
• (2) It is possible that the solution sets are equivalent. One way to check is to substitute values in for most or all parameters (other than k of course) and see if the sets of three numeric series agree. – Daniel Lichtblau Dec 9 '15 at 19:42
• I checked it against direct method. Solution are equivalent, as should be. I also got the results really fast for 4th order equation, which is not the case for direct method. You helped me a lot. – Mikhail Genkin Dec 9 '15 at 20:10
• I still accepted other answer. Even though your answer is usefull, the other one directly answers my question, whether yours is more workaround. Thanks again – Mikhail Genkin Dec 11 '15 at 17:39

You can use the new in M12 function AsymptoticSolve for this:$$\unicode{958}$$

eq = Er k^2 γ ξ (1 + λ Cos[2 ϕ]) (3 I k Cos[ϕ] + 4 σ Cos[2 ϕ] +
I k Cos[3 ϕ]) Subscript[α, ac] + (ξ + σ) (σ + I k Cos[ϕ]) (8 k^4 +
16 k^2 tf1 + 2 k^4 γ + k^4 γ λ^2 + 8 Er k^2 γ σ + 16 Er tf1 γ σ +
4 k^4 γ λ Cos[2 ϕ] + k^4 γ λ^2 Cos[4 ϕ] +
4 k^2 (k^2 + Er γ σ) Sin[2 ϕ]^2 Subscript[μ, 1] +
4 (k^2 + Er γ σ) (k^2 + tf1 - tf1 Cos[2 ϕ]) Subscript[μ, 2]);

AsymptoticSolve[eq == 0, σ, {k, 0, 2}] //TeXForm

$$\left\{\left\{\sigma \to \frac{k^2 \left(3 \gamma \operatorname{Er} \lambda \alpha _{\operatorname{ac}} \cos (2 \phi )+\gamma \operatorname{Er} \lambda \alpha _{\operatorname{ac}} \cos (2 \phi ) \cos (3 \phi ) \sec (\phi )+\gamma \operatorname{Er} \alpha _{\operatorname{ac}} \cos (3 \phi ) \sec (\phi )+3 \gamma \operatorname{Er} \alpha _{\operatorname{ac}}-4 \mu _2 \operatorname{tf1} \cos (2 \phi )+4 \mu _2 \operatorname{tf1}+16 \operatorname{tf1}\right)}{4 \gamma \operatorname{Er} \operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}\right\},\left\{\sigma \to \frac{k^2 \left(4 \lambda \alpha _{\operatorname{ac}} \cos ^2(2 \phi )-3 \lambda \alpha _{\operatorname{ac}} \cos (2 \phi )-\lambda \alpha _{\operatorname{ac}} \cos (3 \phi ) \cos (2 \phi ) \sec (\phi )+4 \alpha _{\operatorname{ac}} \cos (2 \phi )-\alpha _{\operatorname{ac}} \cos (3 \phi ) \sec (\phi )-3 \alpha _{\operatorname{ac}}\right)}{4 \operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}-i k \cos (\phi )\right\},\left\{\sigma \to \frac{k^2 \left(-\lambda \alpha _{\operatorname{ac}} \cos ^2(2 \phi )-\alpha _{\operatorname{ac}} \cos (2 \phi )\right)}{\operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}-\xi \right\}\right\}$$

I tested this result against JungHwan's result for some random values of the parameters, and they agree. The above output is much more concise than JungHwan's results.

Update

Until you have M12, you can use CloudEvaluate:

\$Version
CloudEvaluate[SystemAsymptoticSolve[eq==0, σ, {k, 0, 2}]] //TeXForm
`

"11.3.0 for Mac OS X x86 (64-bit) (March 6, 2018)"

$$\left\{\left\{\sigma \to \frac{k^2 \left(3 \gamma \operatorname{Er} \lambda \alpha _{\operatorname{ac}} \cos (2 \phi )+\gamma \operatorname{Er} \lambda \alpha _{\operatorname{ac}} \cos (2 \phi ) \cos (3 \phi ) \sec (\phi )+\gamma \operatorname{Er} \alpha _{\operatorname{ac}} \cos (3 \phi ) \sec (\phi )+3 \gamma \operatorname{Er} \alpha _{\operatorname{ac}}-4 \mu _2 \operatorname{tf1} \cos (2 \phi )+4 \mu _2 \operatorname{tf1}+16 \operatorname{tf1}\right)}{4 \gamma \operatorname{Er} \operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}\right\},\left\{\sigma \to \frac{k^2 \left(4 \lambda \alpha _{\operatorname{ac}} \cos ^2(2 \phi )-3 \lambda \alpha _{\operatorname{ac}} \cos (2 \phi )-\lambda \alpha _{\operatorname{ac}} \cos (3 \phi ) \cos (2 \phi ) \sec (\phi )+4 \alpha _{\operatorname{ac}} \cos (2 \phi )-\alpha _{\operatorname{ac}} \cos (3 \phi ) \sec (\phi )-3 \alpha _{\operatorname{ac}}\right)}{4 \operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}-i k \cos (\phi )\right\},\left\{\sigma \to \frac{k^2 \left(-\lambda \alpha _{\operatorname{ac}} \cos ^2(2 \phi )-\alpha _{\operatorname{ac}} \cos (2 \phi )\right)}{\operatorname{tf1} \left(\mu _2 \cos (2 \phi )-\mu _2-4\right)}-\xi \right\}\right\}$$


text:
# Estimating effects of a structural break on multiple/panel regressions coefficients and R implementation

I would appreciate some methodology and R implementation help for a thing I'm working on.

I have daily observations of $Y$ for several countries over several years, I will use quite a few independent variables $X$.

My hypothesis is that there was a break at a specific time that has affected the determination process of $Y$ and altered the coefficients of a regression. I’m interested in knowing what the effects of $X$ on $Y$ were before and after the break and whether there was a significant change in these relationships after the break.

My idea is to use multiple regression for each country of the form:

$$Y = a + B_1 X_1 + B_2 X_2 + B_3 X_3 + D + B_4 D X_1 + B_5 D X_2 + B_6 D X_3 + e$$

where $D$ is a dummy variable equal to 1 after the suspected break.

My test for whether the coefficients are different after the break is then simply to test the significance of the coefficients on the interaction terms: $B_4, B_5, B_6$. I know how run this regression for individual countries.

Q1. Will this tell me what I’m looking for or should I use something else like a Wald or Chow test?

Q2. Is this called a natural experiment?

Q3. If I want to run this as a time fixed effects panel regression, how is this done in R?

The suspected break is the introduction date of a new financial regulation. It’s possible that there was a gradual change over a few months in anticipation of the regulation.

Q4. Could I cut out a few months of the data to remove the effects of a gradual change?

Q5. Should I use some method to look for a break before the suspected break date?

Bonus question: In R, the output of these two regressions are the exact same, I’m using the lm function:

$$\begin{array}{rcl} Y & = & a + BX + D + BXD \\Y & = & a + BX + BXD \end{array}$$

Both models return a coefficient specific to the dummy variable D, even though the second model only uses D in the interaction term, why is this and can prevent it?

## 1 Answer

There is a lot going on in the question. You'll get better responses if you narrow the focus of your question. Further, you'll better understand the question yourself.

Here is some suggests which will lead you in the right direction:

# Theory:

### Chow Test

You're going to want to do a simple chow test first around the suspected break data. This is a good start, however, you're example suggests that you'd rather be agnostic about the exact break data, which is a good idea.

### Endogenous Testing

The problem with your suggested approach is that the distribution of the test statistics is not going to follow a typical normal distribution. To begin to understand why this is, image you have 100 different potential breaks and you test each of the resulting 100 dummy variables, using a 5% significant level. By pure luck you're going to fail to reject that dummy variable is different from zero for 5 breaks on average.

So what do you do? Fortunately, some very smart people figured out the correct distribution of the test statistics. The testing procedure is as follows, roll a chow test over your data, compare each of these test statistics to critical values obtained by the aforementioned very smart people. Break date which corresponds to the maximum test statistics, which is statistically significant, is the most likely break date

### Casual Inference

Imagine that you find a break date. Good! One way to estimate the effect of the break is to now run a model which includes a dummy variable which is zero before the date and 1 after. However, the question now is this event casual?

Let's look at a simple example, assume a country deploys some stimulus package on a given date and you want to measure the impact of the stimulus on GDP. Does this coefficient corresponding to the occurrence of the event measure a casual relationship? The answer is, maybe. But probably not. Presumably, the stimulus was dispensed because the country was doing poorly. Therefore, poor GDP could have caused the stimulus, not the other way. This is known as an endogenity problem. The stimulus package is not truly a natural experiment.

# Implementation in R

To implement this in R, you're going to want to use the strucchange package. The documentation is pretty good, here is the vignette

• Great answer, thank you. I read up some more and I have a few quick follow-ups: Is rolling a chow test over the data also called Quandt likelihood ratio/sup-Wald? I guess the function Fstats from strucchange is what I need. In the example in the link you provided they feed an ECM model into this function, do I need that or can I just use a normal lm(Y ~ X)? Could I feed a panel regression into the Fstats function to consider all my countries simultaneously? – Mr. T-stat Mar 25 '17 at 15:59
• I like the point you made about endogeneity problems. Are such issues addressed by theoretical reasoning? I have an argument for why the regulation (break-causer) would affect Y but the other way around doesn’t make that much sense, but I will think about it and include it in my paper. – Mr. T-stat Mar 25 '17 at 15:59
• Yes, you're going to want to use the Fstats. You don't need to feed it any ECM model, any model will work. However, your data many necessitate that you use an ECM model. For example, if your data follows a stochastic trend, (i.e. unit root) then this will impact your testing. You'll want to read about stochastic trends a bit to see if you think you have such a problem. It will depend on you situation and data. – Jacob H Mar 25 '17 at 21:12
• Endogeneity is a very thorny problem and there are not great ways to test if it is present. Generally, the best way to convince people you don't have such a problem is by arguing it. Therefore including an argument within the paper would be a great idea. However, remember, there are other causes of endogeneity. For example, omitted variable bias. – Jacob H Mar 25 '17 at 21:15


text:
# Efficient representation of set of partial order

I guess that notions I describe are already well known, may be by combinatorician, but I do not know their name or any book/article about them. So if you have a link/title I would love to read it.

Let $r$ be an integer, let $P_r$ be the set of partial (pre)order over $[1,r]$ and $T_r$ the set of total (pre)order. I say that $P\in P_r$ is included in $T\in T_r$ if it is included as subset of $[1,r]^2$, or to state it another way, if for all $i,j$ with $i$ less than $j$ for $P$ then it is also less for $T$. Finally I say that $P$ and $P'$ are incompatible if there is $i,j$ with $i<j$ for $P$ and $i>j$ for $P'$ (or if $i=j$ for $P$ and $i<j$ for $P'$).

Let $P\in P_r$ with $i<j$ and no $k$ such that $i<k<j$, then $P_{i,j}$ is the same partial (pre)order, except that $i$ and $j$ are incomparable.

I would like to find an efficient data structure to store a subset of $P_r$. I can't imagine something better than a trie of depth $r(r-1)/2$, with one level for every pair $(i,j)$ with $i<j$. If possible I would want to be able to efficiently add and remove elements from the set, or at least to easily transform $P$ into $P_{i,j}$ as defined above.

I need to know if for a given subset $S$ of $P_r$ and $P\in P_r$, $P$ is incompatible with every $P'\in S$. Or an equivalent problem would be to figure out if a set $S$ is such that its element are one to one incomparable. I also need to know if for every total (pre)order $T$ it is a superset of a partial order $P\in S$.

Intuitively, to each $P\in P_r$ is associate its set $T_P=\{T\in T_r\mid T\supseteq P\}$ and I need to know if $(T_P)_{P\in S}$ form a partition of $T_r$. Then let $T_S=\bigcup_{P\in S} T_P$, then I would also be interested by having a way to efficiently describe $T_S$. (That is, efficiently checking for a given $T\in T_r$ if $T\in T_S$.

Finally, I will introduce the real structure I'm interested in, which is more complicated and probably less usual.

A partial preorder $P$ can be seen as partial function from $[1,r]^2$ to $\{<,>,=\}$, where $i=_P j$ if $i<j$ and $j<i$ in $P$ and $f(i,j)$ is undefined if neither $i<j$ nor $i>j$ in $P$. We can associate to $P$ a subset of $\mathbb N^r$ called $\mathbb N_P$ defined by $(x_1,\dots,x_r)\in \mathbb N_P$ if for all $i<_Pj$ $x_i<x_j$ and for all $i=_Pj$, $x_i=x_j$.

The real data structure I must study is the structure of partial function from $[1,r]^2$ to $\{<,\le,>,\ge,=,\not=\}$. I will call this structure "extended preorder". Let $P$ be an extended preorder, where $(x_1,\dots,x_r)\in \mathbb N_P$ if for all $i<_Pj$ $x_i<x_j$ and for all $i=_Pj$, $x_i=x_j$, for all $i\le j$, $x_i\le x_j$, and for all $i\not_P=j$, $x_i\not=x_j$.

I say that a total preorder $T$ is included in an extended preorder $P$ if $i\le_Pj$ implies $i<_Tj$ or $i=_Tj$, and $i\not=_Pj$ imply $i<_Tj$ or $j<_Ti$.

The operation I must have are still the same, having a set $S$ of extended preorder, verify if an extended preorder is incompatible with every extended preorder of the set, and verify that every total preorder is included in an extended preorder of the set. Or to say it another way, $(\mathbb N_P)_{P\in S}$ is a partition of $\mathbb N^r$.

• Your definition of $P_{i,j}$ seems incomplete. Consider the pre-order $P := \{(1,2),(2,3),(3,4),(4,5)\}$, i.e. in which $1 < 2 < 3 < 4 < 5$. What is $P_{1,5}$? Or do you intend $P_{i,j}$ to be defined only for pairs $i \leqslant j$ for which $\forall k: \Bigl((i \leqslant k) \implies (j \leqslant k)\Bigr) \;\&\; \Bigl((k \leqslant j) \implies (k \leqslant i)\Bigr)$? – Niel de Beaudrap Mar 23 '13 at 14:04
• Indeed it is only defined for those $k$. I edited the question. Thank you – Arthur MILCHIOR Mar 23 '13 at 20:04
• Is the total pre-order problem at the end the one you care more about? If you're actually given a particular total pre-order, that may be easier. (I'll think about the more general problem.) – Niel de Beaudrap Mar 24 '13 at 1:26
• To be more precise, what I really care about is a more complicated structure that may state $i<j$, $i=j$, $i!=j$ or $i<=j$, but I assumed that pre order would be a good first step. Thank you a lot. – Arthur MILCHIOR Mar 24 '13 at 5:33
• Please elaborate — do you mean a data structure which represents not only a pre-order by describing relationships which actually obtain, but also describes invalid ways of extending the order by describing relations which "definitely do not obtain"? – Niel de Beaudrap Mar 24 '13 at 9:47

## Restrictions on pre-orders

You've described that you would like to assert restrictions on a given pre-order: for instance, that specifically $a < b$ rather than merely $a \leqslant b$, so that it would not be compatible with a pre-order in which $a \cong b$ (that is $a \leqslant b$ and $b \leqslant a$). We can achieve this by supplementing the pre-order with a list of forbidden relationships: that is, we specify $a < b$ using a pre-order which specifies $a \leqslant b$ along with many other relationships, and by separately specifying that $b \leqslant a$ is forbidden.

We may similarly represent $a \not\cong b$ by having any pre-order in which at least one of $a \leqslant b$ or $b \leqslant a$ fails to hold. (I got this confused in a previous edit.)

So: we will consider how to represent pre-orders, and presume that we have a list of further restrictions representing relations of the type $\leqslant$, $\geqslant$, and $\cong$ which we forbid between certain pairs. (For each pair $(a,b)$, there are then four possibilities for forbidden relations, including the the case of no forbidden relation.) The forbidden relations I would represent as a list, which we will later iterate through exactly once; however, in order to be able easily to remove restrictions when computing the order $P_{a,b}$ from a pre-order $P$, you might want to combine this list structure with an $r \times r$ array $F$ which stores for each $1 \leqslant a < b \leqslant r$ what sort of relation (if any) is forbidden between $a$ and $b$.

## Representing pre-orders

You can easily represent a partial order by a transitive reduction: a minimal relation (naturally represented as a directed graph, via adjacency lists for immediate predecessors and immediate successors) subject to the constraint of having the partial order as its transitive closure. For a pre-order, you can instead use a relation $R$ which would be the transitive reduction if you "collapsed" all sets of equivalent elements to a single item each; where for all $i \leqslant j$, we have $(i,j) \in R$ if and only if $\exists k: i \leqslant k \leqslant j$ implies that $j \leqslant i$. That is, $(i,j) \in R$ only if one of the following holds:

• $i < j$ and $\neg\exists k: i < k < j$;

• $i \leqslant j$ and $j \leqslant i$.

Such a relation can be easily computed by reduction to computing the transitive reduction of the partial order obtained, as I described, by collapsing equivalence classes (using only one representative node for each equivalence class, and then copying relations across the class). This can be done in time $O(r^3)$ or better.

Given such a representation of pre-orders $P$ by reductions $R$, one may decide $(i,j) \in P$ by testing $(i,j)$-connectivity in $R$, e.g. by breadth-first search. This, of course, takes time $O(n)$. However, one can quite efficiently compute the reduction $R_{i,j}$ of the pre-order $P_{i,j}$ for $i<j$, by

1. copying the immediate successors of $j$ and sharing them with $i$,
2. copying the immediate predecessors of $i$ and sharing them with $j$,
3. removing the relation $(i,j)$ from $R$.

The complexity of this is essentially the sum of the in-degree of $i$ and the out-degree of $j$. I imagine that you would also like to remove the prohibition on $i \leqslant j$ and/or $j \leqslant i$ at the same time; we can achieve this simply by removing any restrictions which are stored at $F[i,j]$ or $F[j,i]$ as appropriate.

## Representing subsets of $S \subseteq P_r$

Assuming that all you want to use the subsets $S \subseteq P_r$ for is to check compatibility of individual pre-orders with the elements of $S$, I think the best approach is to store an $r \times r$ array where each element $(i,j)$ contains a collection which indicates the labels of all pre-orders $P \in S$ such that $(i,j) \in P$. (That is: you do not need to store any forbidden relationships.)

The collection may as well be a simple list, for small sets $S$ or if the orders which it contains don't have too many common relations (e.g. minimal and maximal elements in common); otherwise use your set-implementation of choice (e.g. balanced search trees) instead. This can be updated simply by adding a new relation $P'$ to the entries for all pairs $(i,j)$ which are contained in $P'$. If $P'$ is represented by a reduction $R'$, one may do this by a depth-first search from its minimal elements (which one may store a list of in the representation), and performing depth-first search to find all descendants of each element.

• If one maintains a list of traversed nodes along each path of the traversal, one may add each newly traversed node to the descendants of each of those already on the list; after finishing a traversal, pop the node off the list and mark it as visited.

• For each visited node, we copy its descendants to any nodes on the traversal-list, rather than re-traversing it.

This is also a good thing to do, to compute a more explicit representation of $P'$ from $R'$, once your interest in $P'$ becomes dominated by testing $(i,j) \mathbin{\in?} P'$. This will take time $O(r^2)$, as there is essentially unit cost to reverse each edge $(i,j) \in R$, and to record each of the descendants of any $i \in [1,r]$ throughout the traversal. If $s = |S|$ and you use trees for collections at each index $(i,j)$, constructing the representation for $S$ takes time $O(r^2 s \log s)$, with the $\log s$ factor being saturated in the case where many elements of $S$ share many pairs $(i,j)$ in common. This works best for partial orders which are sparse and essentially unrelated, which would reduce both the expected overlap for pairs $(i,j)$ and the number of entries $(i,j)$ for which any relationship is recorded in the traversal of its reduction.

Having this representation of $S$, you can test incompatibility of $P$ with $S$ (where $P$ is given by a reduction $R$) as follows.

1. Initialise a list of all of the elements of $S$, representing those which are potentially compatible with $P$.

2. Iterate through your forbidden relations, and remove from the list of potentially compatible relations, any element of $S$ which violates any of the constraints. (That is, if $a \leqslant b$ is forbidden, remove any element of $S$ for which that relation holds; and similarly for any element of $S$ fo which $a \cong b$, if that relation is forbidden.)

3. Perform a breadth-first search of $R$: and at each link $(i,j) \in R$ traversed, remove from the list of potentially-compatible orders any order $P$ for which $j < i$, by checking for each order on the list whether this is the case.

If the list of potentially-compatible orders ever becomes empty, then $P$ is incompatible with $S$. Otherwise, the list contains at least one partial order in $S$ with which $P$ is compatible.

We can bound each index search $(i,j)$ for a potentially-compatible order by $O(\log s)$, for $s =|S|$; this will happen for each "reduced" relation in $P$, that is for every ordered pair in $R$. If $m = |R|$, then the worst case is $O(ms \log s) \subseteq O(r^2 s \log s)$, in the case that every element of $S$ is compatible with $P$. If you have $f = |F|$ forbidden relationships, then iterating through these and eliminating potentially compatible relations takes time $O(f \log s)$; this is also dominated by $O(r^2 \log s)$, though the larger $f$ is the faster the subsequent compatibility-checking becomes.

• Thank you a lot for your detailled anwser. There is still one question I don't know the answer of. If for every total order $T$ it is a superset of a partial order $P∈S$. I will also need to figure out if I can extend this to the more complicated data structure I'm really working on. I.e. order stating $i<j$, $i\le j$, $i=j$, $i\not=j$ or $i$ and $j$ may be whatever. – Arthur MILCHIOR Mar 24 '13 at 5:40
• I think I understand: you explicitly meant your remark about partitioning $T_r$, that it is a problem which you would like to efficiently decide. I imagine that this would only happen quite rarely. How large is $S$ typically, and how sparse will its elements be? This problem feels to me like DNF-TAUTOLOGY in that elements of $S$ describe sufficient conditions for a total order to extend it, and you would like to know if all total pre-orders fit one of the sufficient conditions. I may elaborate this answer re: pre-orders, and will await to address your other order type once you've clarified. – Niel de Beaudrap Mar 24 '13 at 10:08
• I am interested in worst case analysis, so I can not tell that $S$ is going to be small or big. I can imagine case where $|S|$ is $1$, or when it is maximal, i.e. ${r!\over 2\log(2)^{r+1}}$. This question happens only one, at the end of the algorithm, I must accept iff $S$ is indeed a partition of $\mathbb N^r$. But it must be efficient, I can't test every total preorder, that's too long. I'll look at DNF-TAUTOLOGY. Thank you a lot – Arthur MILCHIOR Mar 25 '13 at 1:16
• @ArthurMILCHIOR: I have updated my answer, please give it a look. – Niel de Beaudrap Mar 30 '13 at 16:30
• I don't understand why you wrote "we forbid either $a \leqslant b$ or $b \leqslant a$ from holding". Since for $a\not=b$ we would have $a< b$ or $b< a$, which implies $a\le b$ or $b\le a$. Apart from that, it seems helpful, thank you a lot – Arthur MILCHIOR Mar 31 '13 at 8:57


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# Antiparticle for Higgs boson?

Some scientists say that for every particle there exist an anti-particle. If this is true, does it mean that there would be an anti-particle for the Higgs boson also? Could it be that the massless particles would interact with this anti-particle for the Higgs boson?

## 1 Answer

Your first statement is true. However, there is a subtlety - some particles are their own antiparticles.

My understanding is that antiparticles are one particle states (irreducible representations in quantum state space of the Poincaré group see my answer here of what this means) that are mapped into one another by the CPT operator (or better written $C\circ P\circ T$) - time reversal followed by parity inversion followed by charge conjugation.

Some such one particle states are invariant under the action of the CPT operator. It follows by the above definition then that such particles are their own antiparticles. The Higgs Boson is one such particle.

"Antiparticle" and "particle" are partly just words and definitions. Essentially we say that a particle is its own antiparticle if it can emerge alone (i.e. not belonging to a pair) from any quantum state with the requisite momentum, energy and angular momentum. There are no other quantities, such as electric charge, which need to be conserved and therefore pairs of opposite charges or the like are not needed to conserve such quantities.

• Pretty sure you've just defined all SM particles as their own antiparticles. Charge conjugation (not CPT conjugation) is the act of replacing a particle with its antiparticle. The part about 'emerging alone' doesn't make any sense to me. – dukwon Nov 26 '16 at 15:15
• see hyperphysics.phy-astr.gsu.edu/hbase/Particles/antimatter.html "Going beyond the basics, we can say that an antiparticle is related to the particle by charge conjugation. This includes more than just electric charge; it inverts all internal quantum numbers such as baryon number, lepton number, and strangeness. " – anna v Jan 14 '18 at 13:49
• @dukwon I somewhat disagree: the antiparticle is the CPT conjugate, as a property of the Dirac equation. Plain C is violently (maximally) violated in the SM, and P as well, and even T by a little. The vacuum is CPT invariant, and, since the Higgs is shifted by a real number into it in the eponymous mechanism, so is the Higgs itself. The full Higgs complex doublet field, of course, is not. Do you want an answer explicitly demonstrating the field's flip and the conjugate rep gig that achieves that? – Cosmas Zachos Jan 14 '18 at 17:56
• @duckwon. I assume "alone" means "no need for associated production" . So, e.g., in a strong process, $K^0$ cannot emerge "alone", but in association with , e.g., $\bar{K}^0$, because, in this case, strangeness is a conserved quantity "such as electric charge.... or the like", all right. But, of course, it might be for Rod to clarify. My sense is the OP may have developed qualms by the complex Higgs doublet not being its CPT conjugate, but as I indicated, the Higgs and the goldston of $\tau_3$ are self-conjugate. – Cosmas Zachos Jan 15 '18 at 0:51