text:
12th NCERT Matrices Exercise 4.5 Questions 18
Do or do not
There is no try

### Find the adjoint of each of the matrices in Exercise 1 and 2

Question (1)

$\left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ cofactor A
A11 = 4; A12; A21 =2 ; A22 = 1
$\text{cofactor} \;A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\{ - 2}&1\end{array}} \right]$ adj A = (cofactor A)'
$adj\;A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\{ - 2}&1\end{array}} \right]$

Question (2)

$\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\2&3&5\\{ - 2}&0&1\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\2&3&5\\{ - 2}&0&1\end{array}} \right]$ $cofactor\;A = \left[ {\begin{array}{*{20}{c}}3&{ - 12}&6\\1&5&2\\{ - 11}&{ - 1}&{15}\end{array}} \right]$ adj A = (cofactor A)' $adj\;A = \left[ {\begin{array}{*{20}{c}}3&1&{ - 11}\\{ - 12}&5&{ - 1}\\6&2&5\end{array}} \right]$

### Verify A(adj A) = (adj A) A = |A|I in Exercise 3 and 4

Question (3)

$\left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]$ $\left| A \right| = - 12 + 12 = 0$ $cofactor\;A = \left[ {\begin{array}{*{20}{c}}{ - 6}&4\\{ - 3}&2\end{array}} \right]$ adj A = (cofactor A)' $adj\;A = \left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]$ $LHS\; = A\left( {adj\;A} \right)$ $RHS\; = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]$ $RHS\; = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = \left| A \right|I$ $RHS\; = \left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right]$ $RHS\; = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = \left| A \right|I = LHS$

Question (4)

$\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]$ $\left| A \right| = 1\left( {0 - 0} \right) + 1\left( {9 + 2} \right) + 2\left( {0 - 0} \right) = 11$ $cofactorA = \left[ {\begin{array}{*{20}{c}}0&{ - 11}&0\\3&1&{ - 1}\\2&8&3\end{array}} \right]$ adj A = (cofactor A)' $adjA = \left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]$ LHS = A adjA
$LHS = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]$ $LHS = \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]$ $LHS = 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $LHS = 11I = \left| A \right|I$ RHS = adj A. A
$RHS = \left[ {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right]$ $RHS = \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]$ $RHS = 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $RHS = \left| A \right|I$

### Find the inverse of each of the matrices (if it exists) given in Exercise 5 to 11

Question (5)

$\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\4&3\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}2&{ - 2}\\4&3\end{array}} \right]$ $A = 6 + 8 = 14$ $cofactorA = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\2&2\end{array}} \right]$ $adjA = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right]$

Question (6)

$\left[ {\begin{array}{*{20}{c}}{ - 1}&5\\{ - 3}&2\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\{ - 3}&1\end{array}} \right]$ $\left| A \right| = - 2 + 15 = 13$ $cofactorA = \left[ {\begin{array}{*{20}{c}}2&3\\{ - 5}&{ - 1}\end{array}} \right]$ $adjA = \left[ {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right]$

Question (7)

$\left[ {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$ $\left| A \right| = 1\left( {10} \right) - 2\left( 0 \right) + 3\left( 0 \right) = 10$ $cofactorA = \left[ {\begin{array}{*{20}{c}}{10}&0&0\\{ - 10}&5&0\\2&{ - 4}&2\end{array}} \right]$ $adjA = \left[ {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]$

Question (8)

$\left[ {\begin{array}{*{20}{c}}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right]$ $\left| A \right| = 1\left( { - 3 - 0} \right) - 0 + 0 = - 3$ $cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ + 3}&{ - 9}\\0&{ - 1}&{ - 2}\\0&0&3\end{array}} \right]$ $adjA = \left[ {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{ - 3}}\left[ {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right]$

Question (9)

$\left[ {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]$ $\left| A \right| = 2\left( { - 1 - 0} \right) - 1\left( {4 - 0} \right) + 3\left( {8 - 7} \right)$ $\left| A \right| = - 2 - 4 + 3 = - 3$ $cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 4}&1\\5&{23}&{ - 11}\\3&{12}&{ - 6}\end{array}} \right]$ adj A = (cofactor A)'
$adjA = \left[ {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{ - 3}}\left[ {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]$

Question (10)

$\left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]$ $\left| A \right| = 1\left( {8 - 6} \right) + \left( {0 + 9} \right) + \left( {0 - 6} \right)$ $\left| A \right| = 2 + 9 - 12 = - 1$ $cofactorA = \left[ {\begin{array}{*{20}{c}}2&{ - 9}&{ - 6}\\0&{ - 2}&{ - 1}\\{ - 1}&{ + 3}&2\end{array}} \right]$ adj A = (cofactor A)'
$adjA = \left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right]$ ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 2}&0&1\\9&2&{ - 3}\\6&1&{ - 2}\end{array}} \right]$

Question (11)

$\left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]$ $\left| A \right| = - {\cos ^2}\alpha - si{n^2}\alpha = - 1$ $cofactorA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ adj A = (cofactor A)'
$AdjA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]$

Question (12)

Let $A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]$.
Verify that (AB)-1 = B-1 A-1

Solution

$A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]$ $AB = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]$ $AB = \left[ {\begin{array}{*{20}{c}}{18 + 49}&{24 + 63}\\{12 + 35}&{16 + 45}\end{array}} \right]$ $AB = \left[ {\begin{array}{*{20}{c}}{67}&{87}\\{47}&{61}\end{array}} \right]$ $\left| {AB} \right| = 4087 - 4089 = - 2$ $cofactor\;AB = \left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 87}&{67}\end{array}} \right]$ $adj\;AB = \left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 47}&{67}\end{array}} \right]$ ${\left( {AB} \right)^{ - 1}} = \frac{1}{{\left| {AB} \right|}}adj\;AB$ ${\left( {AB} \right)^{ - 1}} = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{61}&{ - 47}\\{ - 47}&{67}\end{array}} \right]$ $A = \left[ {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right]$ $\left| A \right| = 15 - 14 = 1$ $cofactorA = \left[ {\begin{array}{*{20}{c}}5&{ - 2}\\{ - 7}&3\end{array}} \right]$ $adjA = \left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$ ${A^{ - 1}} = \frac{1}{1}\left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]$ ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]$ $B = \left[ {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right]$ $\left| B \right| = 54 - 56 = - 2$ $cofactorB = \left[ {\begin{array}{*{20}{c}}9&{ - 7}\\{ - 8}&6\end{array}} \right]$ $adjB = \left[ {\begin{array}{*{20}{c}}9&{ - 7}\\{ - 8}&6\end{array}} \right]$ ${B^{ - 1}} = \frac{1}{{\left| B \right|}}adjB$ ${B^{ - 1}} = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right]$ $RHS = {B^{ - 1}}{A^{ - 1}}$ $RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right]$ $RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{45 + 16}&{ - 63 - 24}\\{ - 35 - 12}&{49 + 18}\end{array}} \right]$ $RHS = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{61}&{ - 87}\\{ - 47}&{67}\end{array}} \right]$ $RHS = {\left( {AB} \right)^{ - 1}}$ $\therefore {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$

Question (13)

If $A = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]$, show that A2 -5A + 7I = 0
Hence find A-1

Solution

$A = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right]$ $LHS = {A^2} - 5A + 7I$ $LHS = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$ $LHS = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{15}&{10}\\{ - 5}&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right]$ $LHS = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = 0 = RHS$ ${A^2} - 5A + 7I = 0$ ${A^2}{A^{ - 1}} - 5A{A^{ - 1}} + 7I{A^{ - 1}} = 0$ $A - 5I + 7{A^{ - 1}} = 0$ ${A^{ - 1}} = \frac{1}{7}\left[ { - A + 5I} \right]$ ${A^{ - 1}} = \frac{1}{7}\left[ {\left[ {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]} \right]$ ${A^{ - 1}} = \frac{1}{7}\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\1&3\end{array}} \right]$

Question (14)

For the matrix $A = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]$, find the number a and b such that A2 + aA + bI = 0

Solution

$A = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}{11}&8\\4&3\end{array}} \right]$ ${A^2} + aA + bI = 0$ $\left[ {\begin{array}{*{20}{c}}{11}&8\\4&3\end{array}} \right] + a\left[ {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = 0$ $\left[ {\begin{array}{*{20}{c}}{11 + 3a + b}&{8 + 2a}\\{4 + a}&{3 + a + b}\end{array}} \right] = 0$ $\Rightarrow 11 + 3a + b = 0 - - - \left( 1 \right)$ $4 + a = 0$ $a = - 4$ Replace a = -4 in equation(1)
$11 + 3\left( { - 4} \right) + b = 0$ $11 - 12 + b = 0$ $b = 1$ $\therefore a = - 4\; \text{and}\; b = 1$

Question (15)

For the matrix $A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$
Show that A3 - 6A2 + 5A +11I = 0, Hence find A-1

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]$ ${A^3} = \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$ ${A^3} = \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right]$ $LHS = {A^3} - 6{A^2} + 5A + 11I$ $= \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + 5\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] + 11\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $= \left[ {\begin{array}{*{20}{c}}8&{27}&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{24}&{12}&6\\{ - 18}&{48}&{ - 84}\\{42}&{ - 18}&{84}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5&5&5\\5&{10}&{ - 15}\\{10}&{ - 5}&{15}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]$ $= \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0 = RHS$ ${A^3} - 6{A^2} + 5A + 11I = 0$ Multiply by A-1
${A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + 11I{A^{ - 1}} = 0$ ${A^2} - 6A + 5I + 11{A^{ - 1}} = 0$ $11{A^{ - 1}} = - {A^2} + 6A - 5I$ ${A^{ - 1}} = \frac{1}{{11}}\left[ { - {A^2} + 6A - 5I} \right]$ $= \frac{1}{{11}}\left[ { - \left[ {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + 6\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]} \right]$ $= \frac{1}{{11}}\left[ {\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 2}&{ - 1}\\3&{ - 8}&{14}\\{ - 7}&3&{ - 14}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6&6&6\\6&{12}&{ - 18}\\{12}&{ - 6}&{18}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&5\end{array}} \right]} \right]$ $= \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 3}&4&5\\9&{ - 1}&{ - 4}\\5&{ - 3}&9\end{array}} \right]$

Question (16)

If $A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$
Verify that A3 - 6A2 + 9A - 4I = 0, Hence find A-1

Solution

$A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$ ${A^2} = \left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]$ ${A^3} = {A^2}A$ ${A^3} = \left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$ ${A^3} = \left[ {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right]$ $LHS = {A^3} - 6{A^2} + 9A - 4I$ $LHS = \left[ {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $LHS = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0 = RHS$ ${A^3} - 6{A^2} + 9A - 4I = 0$ Multiply by A-1
${A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 9A{A^{ - 1}} - 4I{A^{ - 1}} = 0$ ${A^2} - 6A + 9I - 4{A^{ - 1}} = 0$ ${A^2} - 6A + 9I = 4{A^{ - 1}}$ ${A^{ - 1}} = \frac{1}{4}\left[ {{A^2} - 6A + 9I} \right]$ ${A^{ - 1}} = \frac{1}{4}\left[ {\left[ {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}}9&0&0\\0&9&0\\0&0&9\end{array}} \right]} \right]$ ${A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}3&1&{ - 1}\\1&3&1\\{ - 1}&1&3\end{array}} \right]$

Question (17)

Let A be a nonsingular square matrix of order 3×3. The |adj A| is equal to
(A) |A|   (B) |A|2   (C) |A|3   (D) 3|A|

Solution

${\left[ A \right]_{3 \times 3}}$ $\left| {adjA} \right| = {\left| A \right|^{n - 1}}$ $\left| {adjA} \right| = {\left| A \right|^2}$ Option (B) is correct

Question (18)

If A is an invertible matrix of order 2, then det(A-1) is equal to
(A) det (A)   (B) $\frac{1}{{\det \left( A \right)}}$   (C) 1   (D) 0

Solution

A is an invertible matrix of order 2 $\det \left( {{A^{ - 1}}} \right) = \frac{1}{{\det \left( A \right)}}$ ∴ Option (B) is correct


text:
# PHP5

Subject: PHP5 Package should make clear that PHP5 is required. 2 J Raff 2006-11-13 19:14:54 2006-11-19 14:28:06

 J Raff 2006-11-13 19:14:54 Any package that requires PHP5 or requires another package that requires PHP5 should clearly indicate that it is PHP5 only!

 Connie Müller 2006-11-19 14:28:06 - In reply to message 1 from J Raff you are right... I am tired of that already, as much as I value the efforts of the contributing people as much I am disappointed about this this is not the first script which misses this info this behaviour is not helpful to users, maybe it should be a rule for contributions here

format: plain


text:
## 9.4 Testing

### 9.4.1 Panel or OLS

• It is important to test if the panel regression model is signficantly different from the OLS model. In other words, do we need a panel model or OLS model is good enough?

• The function in the plm package can be used to test a fitted fixed effect model against a fitted OLS model to check which regression is a better choice.

# Simple OLS (without the intercept) using pooling
ols1 = plm(inv ~ value + capital - 1, data = pdata1, model = "pooling")
# summary of results
summary(ols1)
Pooling Model

Call:
plm(formula = inv ~ value + capital - 1, data = pdata1, model = "pooling")

Balanced Panel: n = 10, T = 20, N = 200

Residuals:
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-270.32  -51.32  -23.69  -21.04   -4.51  476.74

Coefficients:
Estimate Std. Error t-value  Pr(>|t|)
value   0.1076384  0.0058256 18.4769 < 2.2e-16 ***
capital 0.1832062  0.0242750  7.5471 1.587e-12 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Total Sum of Squares:    9359900
Residual Sum of Squares: 1935600
R-Squared:      0.8113
F-statistic: 597.659 on 2 and 198 DF, p-value: < 2.22e-16
• The above OLS model can now be tested against the fixed effect model to check for the best fit.
# Testing for the better model, null: OLS is a better
pFtest(fe1, ols1)

F test for individual effects

data:  inv ~ value + capital
F = 50.714, df1 = 10, df2 = 188, p-value < 2.2e-16
alternative hypothesis: significant effects
• Similar to the fixed effect and OLS comparison one can also check if the random effects are needed using one of the available Langrange multiplier tests ( test here) test in function as illustrated below
# plmtest using the Breuch-Pagan method
plmtest(ols1, type = c("bp"))

Lagrange Multiplier Test - (Breusch-Pagan) for balanced panels

data:  inv ~ value + capital - 1
chisq = 727.84, df = 1, p-value < 2.2e-16
alternative hypothesis: significant effects
• A p-value<0.05 in the above test indicates that the Random Effect model is required.

### 9.4.2 Fixed Effect or Random Effect

• The Hausman test () is the standard approach to test for model specification which can be computed using the function in the plm package.
# phtest using the fitted models in fe1 and re1
phtest(fe1, re1)

Hausman Test

data:  inv ~ value + capital
chisq = 2.3304, df = 2, p-value = 0.3119
alternative hypothesis: one model is inconsistent
• A p-value<0.05 suggests that the fixed effect model is appropriate so in this case the random effect model should be used.

### References

Breusch, Trevor Stanley, and Adrian Rodney Pagan. 1980. “The Lagrange Multiplier Test and Its Applications to Model Specification in Econometrics.” The Review of Economic Studies, 239–53.
Hausman, JA. 1978. “Specification Tests in Econometrics.” Econometrica.


text:
## Help with frequency response of guitar tone circuit

The circuit of interest is the basic circuit for a Fender Stratocaster guitar. A diagram can be found at http://www.seymourduncan.com/support...vol_1tone.html . Basically an electric guitar works by vibrating a metal string in a magnetic field. This in turn causes a current to be generated in the an inductor that is wrapped around the magnet.

What I'm trying to do is find the frequency response of this circuit. Whenever I try to calculate the transfer function for this circuit I get

$$TF = 1$$.

When I try to calculate the total impedance for this circuit I get

$$Z_{total} = \frac{Z_{1}Z_{2}}{Z_{1}+Z_{2}} = \frac{(R_{1} - j\frac{1}{\omega C})R_{2}}{R_{1}-j \frac{1}{\omega C}+R_{2}}$$

where
$$Z_{1} = R_{1} - j \omega C$$ and $$Z_{2} = R_{2}$$.

To find the transfer function I calculate

$$TF = \frac{V_{in}}{V_{out}} = 1$$.

This is obviously incorrect, but I don't know where my mistake is. Is my diagram correct?
 PhysOrg.com engineering news on PhysOrg.com >> Army ground combat systems adopts tool for choosing future warfighting vehicles>> Company pioneering new types of material for 3-D printer 'ink'>> Student-built innovations to help improve and save lives
 My post is a litle confusing. I had an ACII diagram, but it did not get rendered the way I wnated it to. R2 is the volume pot, and R1 is the tone pot.


text: I want to prove that $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ using proof by induction. Can science prove things that aren't repeatable? Solve for We,  7.11 = (1-We) * 6 * (1-0.4) + We * 9 ?. Still have questions? 244 0 obj <>/Filter/FlateDecode/ID[]/Index[225 47]/Info 224 0 R/Length 101/Prev 869908/Root 226 0 R/Size 272/Type/XRef/W[1 3 1]>>stream Both and prove it for all A good induction proof is like a can stitch up a conjecture () ≥ where () is the base case. Therefore by the proof of induction it is true for all positive numbers! Divisibility Inductive step: Suppose that 7n-2n is divisible by 5. 2^0 = 1. For n = 1, 7^n - 2^n = 7 - 2 = 5, so the statement is true for n = 1. because 7^n - 2^n is divisible by 5 -- remember, we take the statement as being true, for step 2), 7(7^n - 2^n) = 7(5k) = 35k, therefore this is still divisible by 5, (because we took a number divisible by 5 and multiplied it by another integer), (this is still divisible by 5, since it is the same as above, = 35k), (because 2^n is an integer, and 5 times any integer is divisible by 5), Adding a number divisible by 5, to a number divisible by 5, gives a sum divisible by 5, 7(7^n) - 7(2^n) + 5(2^n) = 35k + 5(2^n) = 5(7k + 2^n), The value on the right of = is divisible by 5 (five times some integer), Therefore, the equal value on the left must be also divisible by 5, Using the statement as being true for n, we proved that the statement remains true for n+1. Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). if you accept the statement being true for n, will it help you prove that it is true for n+1 ? By induction. 5 is divisible by 5. n = 0. k\u\�hڛ�~�����lR�,zu6�vGG�����X��y�5r�. A 2-step proof. %PDF-1.5 %���� 7 - 2 = 5. Use induction to prove that n 3 − 7n + 3, is divisible by 3, for all natural numbers n. Solution : Let P(n) = n 3 – 7n + 3 is divisible by 3, for all natural numbers n. Step 1 : Now P(l): (l) 3 – 7(1) + 3 = -3, which is divisible by 3. End Quiz. Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). As n=k is divisible by 7 then n=k+1 is divisible by 7. Prove true for $n = 1$ Here, both are relatively easy. You remember the "difference of squares"? 17. is divisible by 5. 1 - 1 = 0. . h��W�n7�>&(��ex�-1�@��>l����,Ҧ���g���G����9�9��d���0��P�D�: + n = (1 + 2 + 3 + . Induction basis. That is, 6k+4=5M, where M∈I. 5 is divisible by 5. n = 0. i.e., 7^k - 2^k = 5*m for some integer m, 7^(k+1) - 2^(k+1) = 7*7^k - 2*2^k = 5*7^k + 2*7^k - 2*2^k = 5*7^k + 2*(7^k - 2^k) =, 5*7^k + 2*5*m = 5*(7^k + 2m) which is divisible by 5, thereby completing the proof, 7^n - 2^n = 0 mod 5 : that proves the divisibility by 5. . + 1 (2−1)(2+1) = 2+1 3) 7n – 1 is divisible by 6. 0&0�970�0��>���x��ڼ+o���͝v��� � ĵ���{�����8���H � ��-0 Proof: By induction. 5(7) + . 0 is divisible by any number (because it will always leave a … you may need to re-arrange your equation.) Yes! Therefore, if 7^n-2^n is divisible by 5, so is 7^(n+1) - 2^(n+1). $la�[ ���qБ��d� �Bc The rule generalizes to all powers by saying: In all differences of powers, the difference of roots is a factor: a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3), a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4), (of course, one way to prove this is to use... induction), Once you accept this general rule, it becomes obvious that, a^n - b^n = (a-b)(a polynomial in a and b), 7^n - 2^n = (7-2)(a polynomial in 7 and 2), 5 times an integer is always divisible by 5, (which makes it divisible by any prime factor of 5795). (*****)Now inserting any positive number in always gives a number which is divisible by 7. . 0 is divisible by any number (because it will always leave a … . Attempt. Therefore, by induction, for any positive integer n, 7^n-2^n is divisible by 5. Go through the first two of your three steps: Is the set of integers for n infinite? 225 0 obj <> endobj In a proof by induction that 6n −1 is divisible by 5, which result may occur in the inductive step (let 6k −1 = 5r)? Now finally putting n=1: 12^1 + 2*5^(1-1) = 12 + 2*1 = 14 which is divisible by 7. endstream endobj startxref �1m��dPk����5B���%��bQ���'�w /TPXѢH�2� �K��P"vp ��# � � �[aLd ���{ao�M�]&{��߽��.�O��]=�.���{��KWwMu]ux�H�x$ԖX�-��i'���蟛�v�� Now assume the statement is true for n = k. Our goal is to show that this implies that 7n+1-2n+1 is divisible by 5. 0 ��V %%EOF Prove 6n+4 is divisible by 5 by mathematical induction. 271 0 obj <>stream 7^0 = 1. Here, both are relatively easy. So our property P is: n 3 + 2 n is divisible by 3. We note that 6k+1+4=6×6k+4=6(5M–4)+46k=5M–4by Step 2=30M–20=5(6M−4),which is divisible by 5 Therefore it is true for n=k+1 assuming that it is true for n=k. (why? 60+4=5, which is divisible by 5 Step 2: Assume that it is true for n=k. Get your answers by asking now. Step 3: Show it is true for n=k+1. Checking n=1: 7^1-2^1=5, which is divisible by 5. Step 1: Show it is true for n=0. (N.B. Step 2 : Let us assume that P(n) is true for some natural number n = k. 0 is divisible by any number (because it will always leave a remainder of zero). Find its width.? Since 7-2=5, the theorem holds for n=1. . 7^0 = 1. Doing the two steps proves the statement for all values of n equal to or greater than the value used at step 1. and so on, until everyone tells me to shut up, (and even then, it continues to be true... forever), It should be noted that the statement can be proved using a general rule for "difference of powers". (a) 6k+1 −1 = 5(6r −1), (b) 6k+1 −1 = 6(5r +1), (c) 6k+1 −1 = 5(6r +1), (d) 6k+1 −1 = 5×6r +7. Surgeon general: What to do if you had an unsafe holiday, Report: Sean Connery's cause of death revealed, Padres outfielder sues strip club over stabbing, Biden twists ankle playing with dog, visits doctor, Mysterious metal monolith in Utah desert vanishes, Jolie becomes trending topic after dad's pro-Trump rant, How Biden's plans could affect retirement finances, Legendary names, giant joints and a blueprint for success, Reynolds, Lively donate $500K to charity supporting homeless, Trump slams FBI, DOJ while denying election loss, Wisconsin recount confirms Biden's win over Trump. Suppose that for a positive integer n, 7^n-2^n is divisible by 5. 5) 1 3 3+ 2 2+ 3 + . Mathematical Induction Proof. Hint: It is easy to represent divisibility by$7$in the following way:$8^{n} − 1 = 7 \cdot k$where k is a Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A 2-step proof. h�bbdb� �5 �i;�d{"��e���X��0�L~������"�4�L��$�U.�\$�d�� +n) sturdy sewing machine. endstream endobj 226 0 obj <>/Metadata 21 0 R/Pages 223 0 R/StructTreeRoot 31 0 R/Type/Catalog>> endobj 227 0 obj <>/MediaBox[0 0 612 792]/Parent 223 0 R/Resources<>/Font<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI]/XObject<>>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 228 0 obj <>stream 4) 2n < n! By induction. h�b�Z�af�s|d`��~S;��+C�)f�iW���3�3Q�q������13=�h�������� �9�:���/�����A2�21�5x n = 1. Therefore 6n+4 is always divisible by 5. If the area of a rectangular yard is 140 square feet and its length is 20 feet. Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). n = 1. Here is a more reasonable use of mathematical induction: Show that, given any positive integer n, n 3 + 2 n yields an answer divisible by 3.


text:
## Cryptology ePrint Archive: Report 2020/870

Smoothing Out Binary Linear Codes and Worst-case Sub-exponential Hardness for LPN

Yu Yu and Jiang Zhang

Abstract: Learning parity with noise (LPN) is a notorious (average-case) hard problem that has been well studied in learning theory, coding theory and cryptography since the early 90's. It further inspires the Learning with Errors (LWE) problem [Regev, STOC 2005], which has become one of the central building blocks for post-quantum cryptography and advanced cryptographic primitives such as fully homomorphic encryption [Gentry, STOC 2009]. Unlike LWE whose hardness can be reducible from worst-case lattice problems, no corresponding worst-case hardness results were known for LPN until very recently. At Eurocrypt 2019, Brakerski et al. [BLVW19] established the first feasibility result that the worst-case hardness of nearest codeword problem (NCP) (on balanced linear code) at the extremely low noise rate $\frac{log^2 n}{n}$ implies the quasi-polynomial hardness of LPN at the extremely high noise rate $1/2-1/poly(n)$. It remained open whether a worst-case to average-case reduction can be established for standard (constant-noise) LPN, ideally with sub-exponential hardness.

In this paper, we carry on the worst-case to average-case reduction for LPN [BLVW19]. We first expand the underlying binary linear codes (of the worst-case NCP) to not only the balanced code considered in [BLVW19] but also to another code (in some sense dual to balanced code). At the core of our reduction is a new variant of smoothing lemma (for both binary codes) that circumvents the barriers (inherent in the underlying worst-case randomness extraction) and admits tradeoffs for a wider spectrum of parameter choices. In addition to the worst-case hardness result obtained in [BLVW19], we show that for any constant $0<c<1$ the constant-noise LPN problem is ($T=2^{\Omega(n^{1-c})},\epsilon=2^{-\Omega(n^{min(c,1-c)})},q=2^{\Omega(n^{min(c,1-c)})}$)-hard assuming that the NCP (on either code) at the low-noise rate $\tau=n^{-c}$ is ($T'={2^{\Omega(\tau n)}}$, $\epsilon'={2^{-\Omega(\tau n)}}$,$m={2^{\Omega(\tau n)}}$)-hard in the worst case, where $T$, $\epsilon$, $q$ and $m$ are time complexity, success rate, sample complexity, and codeword length respectively. Moreover, refuting the worst-case hardness assumption would imply arbitrary polynomial speedups over the current state-of-the-art algorithms for solving the NCP (and LPN), which is a win-win result. Unfortunately, public-key encryptions and collision resistant hash functions would need constant-noise LPN with ($T={2^{\omega(\sqrt{n})}}$, $\epsilon'={2^{-\omega(\sqrt{n})}}$,$q={2^{\sqrt{n}}}$)-hardness (Yu et al., CRYPTO 2016 \& ASIACRYPT 2019), which is almost (up to an arbitrary $\omega(1)$ factor in the exponent) what is reducible from the worst-case NCP when $c= 0.5$. We leave it as an open problem whether the gap can be closed or there is a separation in place.

Category / Keywords: foundations / Learning Parity with Noise, Worst-case-to-average-case reduction, Smoothing Lemma


text:
Teruyuki Yorioka is an Associate Professor of mathematics at Shizuoka University.
He mainly studies combinatorics on $\omega_1$ and its related forcing theory.

He received his Ph.D. in 2004 from Kobe University, under the supervision of Joerg Brendle.

When he stayed in Toronto to join the thematic program of set theory 2002, he took the supervision of Stevo Todorcevic.

Personal website

# Recent and upcoming talks by Teruyuki Yorioka

## Independence Results in Mathematics and Challenges in Iterated Forcing, Norwich, November 2-6 2015

Satellite Workshop 2nd November 2015 to 6th November 2015 Held at the University of East Anglia Norwich, UK Organisers: David Aspero (University of East Anglia), Joan Bagaria (Universitat de Barcelona), Mirna Dzamonja (University of East Anglia), Benedikt Loewe (Universität Hamburg) Workshop Theme Independence Results in Mathematics and Challenges in Iterated Forcing Forcing, and especially iterated forcing, is an extremely fruitful technique for proving that certain statements in mathematics are independent from ZFC, or some other base set theory. continue reading…

## Recent Developments in Axiomatic Set Theory, September 16-18, 2015

RIMS Set Theory Workshop 2015 Recent Developments in Axiomatic Set Theory   September 16 (Wed.) 14:20-14:50 Masaru Kada and Takuto Kato: Variants of AC under ZF minus union 15:10-15:40 Masaru Kada and Souji Shizuma: Some remarks on in nite hat guessing games 16:00-16:50 Makoto Takahashi: On non -shortness of Axiom A posets with frame systems September 17  (Thu.) 10:00-10:50 Teruyuki Yorioka: Some consistency results with the existence of a non special Aronszajn tree 11:10-12:00 David Chodounsky: F-Mathias reals and generic filters 14:00-14:50 Joel David Hamkins: Upward closure in the generic multiverse of a countable model of set theory 15:10-16:00 Toshimichi Usuba: Set-theoretic geology and large large cardinals 16:20-17:00 Tadatoshi Miyamoto: Side condition methods and morasses September 18 (Fri.) 9:00-9:50 Sakae Fuchino: On the superuniverse of the set theoretic multiverses 10:10-11:00 Hiroshi Sakai: Covering and approximation properties of ultrapower Venue Room 420, RIMS Kyoto Organizer Masahiro Shioya (University of Tsukuba), E-mail: shioya _at_ math.tsukuba.ac.jp continue reading…

## Infinitary Combinatorics in Set Theory and Its Applications, November 10-13, 2014

RIMS Set Theory Workshop 2014 Infinitary Combinatorics in Set Theory and Its Applications November 10 – 13, 2014 Kyoto, Japan   Overview Infinitary Combinatorics is a classical and main topic of axiomatic set theory. continue reading…

## Sakaé Fuchino conference, November 17-19, 2014

Workshop on Mathematical Logic on the Occasion of Sakaé Fuchino’s 60th Birthday November 17 – 19, 2014 Kobe University, Kobe, Japan   Sakaé Fuchino will be turning 60 this year, and we will celebrate his birthday with a workshop in his honor. continue reading…

## Reflection principles and set theory of large cardinals, September 9 – 12, 2013

The RIMS meeting “Reflection principles and set theory of large cardinals” will take place from Sept. 9 till Sept. 12 at the Research Institute for Mathematical Sciences, Kyoto University, Japan. this meeting is organized by Sakae Fuchino. continue reading…

## Axiomatic approaches to forcing techniques in set theory, November 3-8, 2013

The Banff International Research Station will host the “Axiomatic approaches to forcing techniques in set theory” workshop from November 3rd to November 8th, 2013. The focus of this program is on forcing axioms and their applications within mathematics. continue reading…

## International Conference on Topology and Geometry, September 2-6, 2013

Joint with the 6th Japan-Mexico Topology Symposium, September 2 – 6, 2013 Shimane University and Osaka City University are organizing the international conference on Topology and Geometry 2013. The conference will take place at Shimane University, Matsue City, Japan, on September 2-6, 2013. continue reading…

## 17/Aug/2012: Teruyuki Yorioka, and Dima Sinapova

11:00–12:00 Fields institute, Room 210 Speaker: Teruyuki Yorioka (Shizuoka) Title: An application of Aspero-Mota’s iteration to Todorcevic’s OCA Abstract: It is an open question whether it is consistent that Todorcevic’s OCA holds together with the continuum larger than $\aleph_2$. continue reading…

## Forcing extensions and large cardinals, December 4 – 7, 2012

Dates, venue, and organizer Dates: December 4 (Tue) – 7 (Fri), 2012 Venue: Room 111, Research Institute for Mathematical Sciences, Kyoto University, Kyoto, Japan Organizer: Tadatoshi Miyamoto (Nanzan University). Overview By strengthening the standard Zermelo-Fraenkel axiom system of set theory (ZFC), one can decide a number of important statements in mathematics, like e.g. continue reading…


text:
, 03.07.2019 19:30 toxsicity

# Acarpenter had a board that was 5.28 m5.28 \text { m} 5.28 m 5, point, 28, space, m long. she used her circular saw to cut 9 cm9 \text{ cm} 9 cm 9, space, c, m off the end of the board. how many centimeters long was the board after the carpenter cut it?

### Another question on Mathematics

Mathematics, 21.06.2019 21:00
Abakery recorded the number of muffins and bagels it sold for a seven day period. for the data presented, what does the value of 51 summarize? a) mean of bagels b) mean of muffins c) range of bagels d) range of muffins sample # 1 2 3 4 5 6 7 muffins 61 20 32 58 62 61 56 bagels 34 45 43 42 46 72 75
At the county fair, the baxter family bought 6 hot dogs and 4 juice drinks for $12.90. the farley family bought 3 hot dogs and 4 juice drinks for$8.55. find the price of a hot dog and the price of a juice drink.


text:
Register Forum Rules FAQ Members List Social Groups Search Today's Posts Mark Forums Read

 AnimeSuki Forum MKV to MP4 Guide
 User Name Remember Me? Password

 Notices Trouble logging in?If you can't remember your password or are having trouble logging in, you will have to reset your password. If you have trouble resetting your password (for example, if you lost access to the original email address), please do not start posting with a new account, as this is against the forum rules. If you create a temporary account, please contact us right away via Forum Support, and send us any information you can about your original account, such as the account name and any email address that may have been associated with it. There are changes planned for the forum that are known to be incompatible with older browsers. In particular if you are still using Internet Explorer on Windows XP (or similar old OS or browser), please consider upgrading a.s.a.p. or the forum will stop loading for you. Detailed information can be found in this thread.

 2007-05-17, 10:15 Link #1 martino makes no files now     Join Date: May 2006 MKV to MP4 Guide Recently there has been an upsurge of people asking how to convert MKV to MP4 so that they can play it on their XBOX360 or PS3 (You have to note that the PS3 does not support the High profile yet, in which most fansubs are encoded, so it's very likely that those files will not play unless they are re-encoded to the supported profile). So I have decided to put something like this us since it will be easier than having dozens of threads which try to achieve the same purpose. Also, scroll down to the very bottom to read a bit more... Before you even start... Spoiler for boring stuff, read only if you actually want to know what you are getting yourself into: What files do I need? Firstly you need to grab the script itself from here. Save it as mkv2mp4.bat and it's version 1.03 at the moment. The next thing you need to do is unpack the bat script into some folder. Then create a new folder called "bin" inside it. Also make sure to have CCCP installed and fully working. Now, the biggest requirement for this to work is to have mkvtoolnix, which is a tool that allows tracks to be removed from MKV files and put back in. You can download it from here. Install it as usual. Then navigate to the directory where you installed it and copy the following files into the bin directory that you created earlier on. Spoiler for list of files that need to be copied: You need to download MP4Box which just simply muxes streams into the MP4 container, which is supported by both the PS3 and XBOX360. You can get it from here and copy the executable (and any *.dll file, if present) into the bin directory. You will also need a little package that will do some pretty smart "magic". It's this what makes this whole remuxing process possible without having to add too much user required input. Get it from here and put all of its contents into the bin directory (if it requests a password for the package then use doom9.org). You need to download and install AviSynth 2.5 for video frameserving since you might be re-encoding the video. You can get it from here. You need to download the nero AAC encoder from here and place neroAacEnc.exe into the bin directory. Same with BePipe, placing bepipe.exe into the bin directory. Furthermore you will need to download x264, which is the H.264/AVC1 encoder application that might be used. It can be found here. The topmost *.7z file is what you'll be usually looking at getting and put x264.exe into the bin directory. Lastly, you need a little tool that does some hex changing, I won't go into details, but it is needed because both Sony and Microsoft fail pretty hard. Get it from here and again place the executable into the bin directory. All these tools should suffice for all the tasks in this script. What do I do next then? The batch script executes different commands based on what media file you have, since not all of them can be treated the same. In this section I will explain how to identify them. The simplest approach would be to open the media file in Media Player Classic, right-click on the video and move to the Details tab. There you will see a basic description of the file. Type 1 Video: MPEG4 Video (H264) Audio: AAC - just remuxing needed Type 2 Video: MPEG4 Video (H264) Audio: MPEG Audio Layer 3 (aka MP3), Vorbis, AC3 or any others - audio transcoding to AAC needed, rest will be just remuxed Type 3a Video: XVID, DIVX or any others Audio: AAC - video transcoding needed, audio will be remuxed Type 3b Video: MPEG4 Video (H264), XVID, DIVX or any others Audio: AAC Subtitle: Advanced SubStation Alpha or any others - video transcoding needed while hardsubbing subtitles, audio will be remuxed Type 4a Video: XVID, DIVX or any others Audio: MPEG Audio Layer 3 (aka MP3), Vorbis, AC3 or any others - video and audio transcoding needed Type 4b Video: MPEG4 Video (H264), XVID, DIVX or any others Audio: MPEG Audio Layer 3 (aka MP3), Vorbis, AC3 or any others Subtitle: Advanced SubStation Alpha or any others - video and audio transcoding needed while hardsubbing subtitles About the pixel aspect ratio Since version 1.02 I had to introduce an option to specify the pixel aspect ratio in order to retain the correct look of the video. Unfortunately this currently cannot be automated and has to be set by the user. First thing to do is to open once again the basic properties of a file in Media Player Classic as described in the previous section and look closely at the video part. If you see something like "Video: MPEG4 Video (H264) 704x480 (853:480) 23.98fps [Video]" or "Video: MPEG4 Video (H264) 720x480 (853:480) 23.98fps [Video]" or even "Video: MPEG4 Video (H264) 704x400 (711:400) 23.98fps [Video]" it means that there was a custom ratio set by the person who made the video file (for those interested the first resolution is the one at which the file was encoded and the second one, in brackets, is the one that it gets stretched to on playback). You need to make sure to do the same. If there is just one resolution listed it means that nothing was set and you should just choose "1:1" when the option appears. I'll just simply list the most common possibilities and their respective pixel aspect ratios for ease of use. 704x480 (853:480) or 704x480 (16:9) - 40:33 720x480 (853:480) or 720x480 (16:9) - 32:27 704x400 (711:400) or 704x400 (16:9) - 100:99 704x480 (640:480) or 704x480 (4:3) - 10:11 720x480 (640:480) or 720x480 (4:3) - 8:9 For other non-common ones please contact me and I'll try to help you. Getting it all down Before starting the script, make sure to move the file that you would like to convert into the directory where you unpacked the mkv2mp4.bat script and where the bin subdirectory is located. Now run the script by double-clicking on it. Firstly it will ask for the type of file that you would like to convert. You should have already determined that from the previous section. The possible choices are from 1 to 4. A word of warning before I proceed to the next step. Please make sure that the files that you want to convert don't have any spaces in them. Best thing is to just replace them with underscores. Same goes for any other files that you may be using later on. Depending on your choice it will ask for other inputs which I'll describe next; Type 1 1) This one will ask next for the input filename of the file that you would like to convert without the extension. 2) Then it will ask for the output file, without the extension again, which is just the name of the file that it will create using the MP4 container. After that it should just work through all the lines in the script, delete all the intermediate files at the end and place the output into the directory where the script is found. And you're done. Type 2 1) Same as step 1 and 2 from Type 1 Type 3 1) Firstly you need to change one setting in CCCP. Go into its Settings application, click Next and check "Autoload VSFilter" and Apply. 2) You need to create an AviSynth script. Place the MKV file into the directory where you unpacked the mkv2mp4.bat script, and create the script there as well. Here is a basic example of an AviSynth script; Code: DirectShowSource("C:\path\to\file.mkv") 3) Same as step 1 in Type 1 4) It will ask for the AviSynth script filename, without the extension. 5) Same as step 2 in Type 1 6) Now it will ask whether the video is standard or high definition. This just tells the script to use different encoding settings, since for once you don't really want to kill the encoding speed or playback too much, right? 7) Finally it will ask for the bitrate. Usually go with some value given in the recommendations range. It should give decent results. 8) Now just wait for it to do its job and spit out an MP4 video file. Type 4 1) Same as steps 1-8 in Type 3 --- Spoiler for the old "manual" way (no longer updated): Constructive feedback is of course welcome. (Hopefully I haven't missed anything important or made some silly mistake, since you could consider this as an intermediate version.) __________________ "Light and shadow don't battle each other, because they're two sides of the same coin" Last edited by martino; 2008-12-19 at 12:26. Reason: v1.03 update
2007-05-17, 13:07   Link #2
GHDpro
Administrator
Administrator

Join Date: Jan 2001
Location: Netherlands
Age: 36
Alternative Guide
This version is a bit easier. However, as the video will be re-encoded (even if the subtitles are already encoded into the video), the result will most likely be lower quality.

1. Required Software
- MKVExtractGUI
- mkvtoolnix
- VSFilter (part of CCCP, so you should have it already)
- PS3 Video 9 (similar tools might also work for other systems such as PSP or Wii)
- AviSynth (optional - also included in PS3 Video 9!)

The first three files are only necessary if the file you are trying to convert MKV files with soft subtitles.

2. Extracting the subtitles
If the file has soft subtitles, you need to extract the subtitles first using MKVExtractGUI. Make sure you put MKVExtractGUI in the same folder as mkvtoolnix (C:\Program Files\mkvtoolnix by default) btw.

The subtitle file (shown as S_TEXT/ASS for styled subtitles or S_TEXT/UTF8 for plain subtitles) will be put in the same folder as the video file.

Note: some MKV files also include fonts. You may need to extract and install the fonts (using the Control Panel) in order for the styled subtitles to look just the way they're supposed to look.

3. Write the AVS (AviSynth) file
This is the hard part: writing the AVS file. Actually the hard part is getting the folders and filenames just right (this could be tricky if the file has a complicated filename and is in a folder a few levels deep). Btw, an AVS file is simply a text file with a .AVS file extension.

Write a AVS file that looks like this:
Quote:
 LoadPlugin("C:\Program Files\Combined Community Codec Pack\Filters\vsfilter.dll") DirectShowSource("C:\Anime\Hayate no Gotoku\[SS-Eclipse] Hayate no Gotoku! - 07 (1280x720 h264) [367A4C60].mkv") TextSub("C:\Anime\Hayate no Gotoku\[SS-Eclipse] Hayate no Gotoku! - 07 (1280x720 h264) [367A4C60]_Track3.ass")
Of course make sure all the paths and filenames are changed to whatever file you want to convert. TIP: right click and select "Rename" on the file you want the filename of, and then press CTRL-C to copy the filename to the clipboard. Then just press CTRL-V to paste the filename in Notepad.

If the MKV file you want to convert does not contain subtitles (in case they're already encoded in the video) leave out line 1 ("LoadPlugin") and line 3 ("TextSub") and keep just line 2 ("DirectShowSource").

Before converting, verify the AVS file works by opening it in MPC (Media Player Classic). If you get a AviSynth error message, you'll need to fix the AVS file.

4. Convert with PS3 Video 9
I'm not going to explain the usage of PS3 Video 9 here, as it's pretty simple. Also you'll need to test for yourself which conversion profile works best for you. But if you have followed the steps above, you should be able to convert the MKV file just like you would any AVI file.

If you need a hint: select "Convert", then "Current conversion" tab. Then just click the "Convert video" button and select the AVS file you just created.

Optional hint
If you go to Start -> Programs -> Combined Community Codec Pack -> Media Splitter Settings and then under the Options tab enable "Autoload VSFilter" (in Compatibility section), you might be able to skip the whole MKVExtractGUI phase! You probably still need to write the AVS file, but you can get away with just the "DirectShowSource" line (and leave out the others).

Disclaimer
I have neither a PS3 nor a Xbox 360, so I can't test this. But I did succesfully convert various files to PSP format (using PSP Video 9) this way.

Last edited by GHDpro; 2007-05-17 at 14:50.

 2007-05-17, 19:39 Link #3 Nicholi King of Hosers   Join Date: Dec 2005 Age: 33 Loading VSFilter within DirectShowSource (if you use Autoload VSFilter) might get flipped images and such depending, just be sure to preview before you hit encode. Also specific note in terms of re-encoding while using the CCCP's ffdshow, you should turn on YV12 ouput in FFDShow Video Decoder Config under Outputs. Less colorspace conversions = more good for you. Next CCCP version will already have this on by default, but currently all past versions have used YUY2.
 2007-05-18, 17:45 Link #4 Aegelward Junior Member   Join Date: Feb 2006 i've figured out a way to do this on the mac, it eats up a bit of disc space however. and it only works for files with one audio track. doing converts straight from viddyup sometimes resulted in stuttering files so... make sure you have perian 1.0 (beta) installed (http://perian.org/beta.html) and download viddy up (www.splasm.com) the unregistered version works fine, as we are only going to transfer it into a .mov (renaming won't work) from there you can use isquint (www.isquint.org) or visualhub (£££) (http://www.techspansion.com/visualhub/) ... well, personally i just use the apple tv settings & h.264 - high or standard. some files i notice a noticable quality difference, some i don't. just depends on the original encode i guess. - oh, directly converting from mkv using visualhub sometimes works, but this is a surefire way to make sure you get subtitles.
 2007-08-10, 14:25 Link #5 Hinosaki Junior Member   Join Date: Aug 2007 Hmm, I have a slight problem. I successfully converted one file to .MP4 and it played fine. However when I was working on my second one I get this problem. Spoiler for : Not sure what to do here. I rechecked the steps and redid everything but the same problem occurs. EDIT : Scratch that. I got it to work somehow. I guess I mistyped something. >_>. Last edited by Hinosaki; 2007-08-10 at 15:17. Reason: Figured it out.
 2007-08-11, 21:58 Link #6 kujoe from head to heel     Join Date: Jan 2004 Location: Vancouver, Canada Age: 33 Ok, I'm obviously a newbie at this... so, I have question if you guys don't mind me asking. Do any of the illustrated methods work for videos (.mkv) with multiple sound channels? I would like to convert something which has "AAC 48000Hz 6ch [Audio]." EDIT: Never mind. Figured it out for myself. Thanks for the guides. Last edited by kujoe; 2007-08-13 at 20:26.
 2007-12-01, 14:41 Link #7 Fleabitten8907 Loves stripes.   Join Date: Sep 2006 Age: 26 Question/concern~ So like, I'm using the alternate method since it looked more nub-friendly, and for some reason I'm having problems. When I get to the conversion process, it for some reason won't add to my iTunes library, and when I try to open it manually, it will only play audio in Winamp. However, using the exact same method, but using the 30second test option, it works perfectly.... What could be causing this? x_x
 2007-12-01, 21:10 Link #8 hobbes_fan You could say.....     Join Date: Apr 2007 I've put a link to this in my sig. I also request that this be stickied. It has become a once a week question. __________________ Before you ask "How do I convert fansubs to...." see the following MP4 - http://forums.animesuki.com/showthread.php?t=47693 Convert AVI/MKV/MP4 to DVD http://forums.animesuki.com/showthread.php?t=26308
2007-12-01, 21:40   Link #9
martino
makes no files now

Join Date: May 2006
Quote:
 Originally Posted by Fleabitten8907 Question/concern~ So like, I'm using the alternate method since it looked more nub-friendly, and for some reason I'm having problems. When I get to the conversion process, it for some reason won't add to my iTunes library, and when I try to open it manually, it will only play audio in Winamp. However, using the exact same method, but using the 30second test option, it works perfectly.... What could be causing this? x_x
You shouldn't be really using Winamp to play the output file. It's meant to be for audio playback and its handling of video files just simply doesn't cut it. Not sure what iTunes has to do with this, maybe it's somehow interfaced with that xx Video 9 program. I've never run that program so I wouldn't know, sorry.

Also as an update I will attempt to write a script for ease of use, after my exams though. However I currently do have a batch script for converting any file that plays in DirectShow to XviD/MP3 in AVI. It's currently being tested... Just a heads up.
__________________
"Light and shadow don't battle each other, because they're two sides of the same coin"

 2007-12-21, 07:08 Link #10 zuka Junior Member     Join Date: Jul 2007 Location: Canada mp4mobile is probably the best choice if you are willing to pay for it http://www.videomobileconverter.com/mp4Mobile/
 2008-01-07, 05:55 Link #11 DuplicateGoose Junior Member     Join Date: Jan 2008 Location: Nottinghamshire, England Age: 25 Metroska (.mkv etc) to common Windows video file types I'm not sure I want to go into great detail explaining this but, I always managed to convert Metroska file containers to more common video file types like .wmv or MPEG2 using the Combined Community Codec Pack and the full version of TMPGEnc 4.0 Xpress. I'm new to the forum and joined because I found a great bunch of technical support threads through Google. I hope this has helped anybody who was hoping in not having to overheat their heads over the challenge and I'm sure you can pick up TMPGEnc Xpress through torrent searching.
 2008-01-14, 20:15 Link #12 hobbes_fan You could say.....     Join Date: Apr 2007 you realise that this product is a retail product. TMPENGC v4.0 is not a free product and any torrented version of it is pirated. It is sold for \$110USD http://forums.animesuki.com/faq.php?...#faq_rules_1_1 __________________ Before you ask "How do I convert fansubs to...." see the following MP4 - http://forums.animesuki.com/showthread.php?t=47693 Convert AVI/MKV/MP4 to DVD http://forums.animesuki.com/showthread.php?t=26308
 2008-01-14, 21:11 Link #13 Quarkboy Anime Translator     Join Date: Nov 2003 Location: Tokyo, Japan Age: 36 I just noticed this thread... and came to gloat a bit. Ah, I remember the days back in 2005 when I was ridiculed for switching to h.264 in mp4 in my group (one of, if not the first group to do so)... And now to see people actually requesting their mkvs be converted into mp4s, for exactly the reason I predicted (better hardware playback compatability) makes my day. I love being right. __________________ Yomiuri Television Enterprise International Media Strategy Chief Sam Pinansky
 2008-01-21, 23:38 Link #14 garra113 CaramellDansen!!!   Join Date: Apr 2006 I have a question about this. Whenever I encode mkv to mp4 files, it will leave me too extract the fansubs and merge them back to the mp4 file. Question #1: Is it possible for a way to merge the Typesetting along with the subtitles into the newly encoded mp4 file? Question #2: If it don't work, then is there like another way for it to see it with typesetting using the fansub group typesetting?
 2008-01-22, 10:30 Link #15 Salt the cynic     Join Date: Nov 2006 Location: Somewhere between life and death. Age: 34 If you are just converting for your PSP then there's also XviD4PSP, which contrarily to it's name, handles just about any file format. It seems to work OK. It appears to uses FFdshow, so it your mkv has subtitles, you might want to go to FFdshow video decoder options to turn on subtitles and embedded subtitles. The subtitles will end up hardcoded to the resulting MP4 file. __________________ De gustibus non est disputandum.
 2008-01-22, 12:09 Link #16 garra113 CaramellDansen!!!   Join Date: Apr 2006 Yeah but the whole point of encoding it to mp4 is for the softsubs though. Anyway to do it the way I said up my 1st post?
2008-01-22, 17:51   Link #17
martino
makes no files now

Join Date: May 2006
MP4 does not support softsubs, ok maybe it does, but no device supports the TTXT format. Plus, you're not gaining anything by converting a softsubbed MKV to a softsubbed MP4, in terms of niceness and compatibility. It's just... pointless.

Quote:
 Originally Posted by Salt It appears to uses FFdshow, so it your mkv has subtitles, you might want to go to FFdshow video decoder options to turn on subtitles and embedded subtitles. The subtitles will end up hardcoded to the resulting MP4 file.
Hopefully not. It should be using VSFilter's rendering, and not FFDShow's which should be used just as the audio/video decoder. That is if it is using DirectShow output for frameserving.

Quote:
 Originally Posted by garra113 I have a question about this. Whenever I encode mkv to mp4 files, it will leave me too extract the fansubs and merge them back to the mp4 file. Question #1: Is it possible for a way to merge the Typesetting along with the subtitles into the newly encoded mp4 file? Question #2: If it don't work, then is there like another way for it to see it with typesetting using the fansub group typesetting?
Merge? What do you mean by that... Is that like hadrsubbing it, or converting to TTXT?
__________________
"Light and shadow don't battle each other, because they're two sides of the same coin"

 2008-01-22, 18:42 Link #18 garra113 CaramellDansen!!!   Join Date: Apr 2006 I wish to see the same typesetting as the fansub groups already provided in the mp4 BUT the mkvmerge doesn't allow you to merge the typesettings and only the subtitles. In other words, I wish to merge the typesettings but I don't know how loll.
2008-01-25, 09:46   Link #19
Salt
the cynic

Join Date: Nov 2006
Location: Somewhere between life and death.
Age: 34
Quote:
 Originally Posted by martino Hopefully not. It should be using VSFilter's rendering, and not FFDShow's which should be used just as the audio/video decoder. That is if it is using DirectShow output for frameserving.
LOL. I have no idea what you are talking about.

I'm clueless as to what goes on behind the scenes (hence my using of a program with a relative simple GUI).

I had to fiddle around quite a bit to get subtitles to appear (hardcoded of course) in the resulting PSP compatible MP4 file. I tried adjusting DirectVobSub's setting, to no valid, then I read something about turning on the subtitles in FFDshow, and that did the trick.
__________________
De gustibus non est disputandum.

2008-01-25, 12:19   Link #20
martino
makes no files now

Join Date: May 2006
Quote:
 Originally Posted by garra113 I wish to see the same typesetting as the fansub groups already provided in the mp4 BUT the mkvmerge doesn't allow you to merge the typesettings and only the subtitles. In other words, I wish to merge the typesettings but I don't know how loll.
Typesetting is usually hadsubbed on the video, or softsubbed in the same track as the dialogue because you can't have two softsub tracks playing at the same time. Also, what are you trying to achieve? Softsub an MP4? For what point? It's totally pointless as I posted before...
__________________
"Light and shadow don't battle each other, because they're two sides of the same coin"

 Tags wiki candidate, tutorial, file conversion

 Thread Tools

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Anime Discussion     Current Series         Durarara!!         Fairy Tail         GATE         Gundam         Heavy Object         Monogatari Series         Muv-Luv Franchise             Total Eclipse         Naruto         One Piece         Utawarerumono     Older Series         Accel World         Aldnoah.Zero         Angel Beats!         Aquarion Series         Attack on Titan         Bleach         Chuunibyou         Code Geass         Fate/ Series             F/SN UBW TV Spoiler Threads (for Franchise Veterans)             F/SN UBW TV Spoiler-Free Threads (for New Viewers)         Haruhi Suzumiya         Hidan no Aria         Hayate no Gotoku         High School DxD         Infinite Stratos         KanColle         Kill La Kill         Little Busters         Log Horizon         Macross             Past Macross Series         Madoka Magica         Mahouka         Nanoha         No Game No Life         Psycho-Pass         Soul Eater         Steins;Gate         Suisei no Gargantia         Sword Art Online         To Aru Kagaku no Railgun         To Aru Majutsu no Index         Valvrave the Liberator         Retired             Retired A-L                 Ah! My Goddess                 Air                 Amagami SS                 AnoHana                 Another                 Asatte no Houkou                 Chrno Crusade                 Clannad                 Boku wa Tomodachi...                 Claymore                 Danganronpa                 Darker Than Black                 Death Note                 Eden of the East                 Full Metal Panic                 Fullmetal Alchemist                 Guilty Crown                 Gurren-Lagann                 Hanasaku Iroha                 Higurashi                 Horizon                 Hyouka                 K-On!                 Kanon                 Kashimashi                 Kami nomi/TWGOK                 Kimi ga Nozomu Eien                 KonoMini                 Kannagi                 Lucky Star             Retired M-Z                 Maria-sama                 Midori no Hibi                 Monster                 My HiME / My Otome                 Myself; Yourself                 Negima                 Ouran Host Club                 Ore no Imouto                 Penguin Drum                 Pumpkin Scissors                 Regios                 Rozen Maiden                 School Days                 School Rumble                 Sakurasou                 Shana                 Shuffle!                 sola                 Sora Kake                 Shin Sekai Yori                 Spice and Wolf                 Tales of...                 Toradora!                 True Tears                 Umineko                 Valkyria Chronicles                 Vampire Knight                 Zero no Tsukaima     Upcoming Series         Upcoming Archive     Movies, OVAs/OADs, and Specials Anime Related Topics     Light Novels         High School DxD [LN/M]         Mahouka [LN/M]         To Aru... Index [LN/M]     Web Novels     Manga         To Aru... Railgun [M/LN]     Visual Novels, Mobage & Anime Spin-Off Games     Music     Fan Creations     Reviews & Recommendations         Reviews         Custom Requests     General Anime         Fansub Groups     Archived Manga & Light Novel Discussion General     General Chat     News & Politics     Sports & Entertainment     Science & Technology     Video Games Support     Tech Support         Playback Help     Forum & Site Feedback

All times are GMT -5. The time now is 03:06.

 -- vBulletin -- AnimeSuki Default -- AnimeSuki Default (Contacts Highlighted) -- AnimeSuki Default (Original Navigation) -- AnimeSuki Compressed -- AnimeSuki Blue -- AnimeSuki Blue (Reordered) Contact Us - Support - AnimeSuki.com - Archive - Top

Powered by vBulletin® Version 3.8.9
Copyright ©2000 - 2016, vBulletin Solutions, Inc.
We use Silk.


text:
×

# Integration Challenge 3! [Courtesy: Hasan Kassim]

$\large \displaystyle \int_{0}^{\infty}{\frac{{x}^{a}}{1+{x}^{n}} \ln\left(\frac{1+{x}^{n}}{{x}^{m}}\right) \ dx} \ = \ \large \frac{\pi}{n} \csc\left(\frac{\pi(a+1)}{n}\right) \left(\frac{m\pi}{n}\cot\left(\frac{\pi(a+1)}{n}\right) - \left(\psi\left(\frac{n-a-1}{n}\right) + \gamma\right)\right)$

Prove the identity above.

Note by Kartik Sharma
2 years, 1 month ago

Sort by:

Consider $$\displaystyle\text{I}(z) = \int_{0}^{\infty}\dfrac{x^a}{(x^n+1)^z} \mathrm{d}x$$

Putting $$x^n+1=\dfrac{1}{t}$$, we have,

$$\displaystyle\text{I}(z) = \dfrac{1}{n} \int_{0}^{1} t^{z-\left(\frac{a+1}{n}-1\right)}(1-t)^{\left(\frac{a+1}{n}-1\right)} \mathrm{d}t$$

$$\displaystyle = \dfrac{1}{n} \operatorname{B}\left(\dfrac{a+1}{n}, z-\dfrac{a+1}{n}\right)$$

$$\displaystyle=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma\left(z-\dfrac{a+1}{n}\right)}{\Gamma(z)}$$

$$\displaystyle\implies \text{I} '(z)=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma ' \left(z-\dfrac{a+1}{n}\right)\Gamma(z)-\Gamma '(z)\Gamma\left(\dfrac{a+1}{n}\right)\Gamma\left(z-\dfrac{a+1}{n}\right)}{(\Gamma (z))^2}$$

$$\displaystyle\implies \text{I} '(1)=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma ' \left(1-\dfrac{a+1}{n}\right)\Gamma(1)-\Gamma '(1)\Gamma\left(\dfrac{a+1}{n}\right)\Gamma\left(1-\dfrac{a+1}{n}\right)}{(\Gamma (1))^2}$$

Now, since $$\displaystyle\Gamma '(z)=\psi(z)\cdot\Gamma(z)$$ and $$\Gamma '(1)=-\gamma$$ and using Euler's reflection formula, we have,

$$\displaystyle\text{I}'(1)=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right)$$

Also, from the original integral,

$$\displaystyle\text{I}'(1)=-\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x$$

$$\displaystyle\therefore \int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x = -\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right) \ (1)$$

Next, let $$\displaystyle\text{J}(a)= \int_{0}^{\infty}\dfrac{x^a}{(x^n+1)}\mathrm{d}x$$

Note that $$\text{J}(a)=\text{I}(1)$$

$$\displaystyle=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma\left(1-\dfrac{a+1}{n}\right)}{\Gamma(1)}$$

$$\displaystyle=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\right)$$

Therefore,

$$\displaystyle\text{J}'(1)=\int_{0}^{\infty}\dfrac{x^a}{1+x^n}\ln x \ \mathrm{d}x = -\dfrac{{\pi}^{2}}{n^2}\csc\left(\dfrac{\pi(a+1)}{n}\right)\cot\left(\dfrac{\pi(a+1)}{n}\right) \ (2)$$

Operating $$(1)-m\times(2)$$, we have,

$$\displaystyle\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln \left(\dfrac{x^n+1}{x^m}\right)\mathrm{d}x=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\right)\left[\dfrac{m\pi}{n}\cot\left(\dfrac{\pi(a+1)}{n}\right)-\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right]$$

Q.E.D.

- 2 years, 1 month ago

@Kartik Sharma Can you please re-check whether R.H.S. contains $$+\gamma$$ or $$-\gamma$$ in the bracket?

- 2 years, 1 month ago

Edited! Great! Congratulations once again! (You can post the proof of the first challenge even though I've done already, a new proof will be accepted; just in case I thought you must be interested).

- 2 years, 1 month ago

so LoveLy As always

- 2 years, 1 month ago

I'm not so good at high level integrals. But I'm writing all the proofs provided by u and trying to use them. Thanks a lot man. Keep on posting proofs!

- 2 years, 1 month ago


text:
## moneybird Group Title Compute $\int\limits_{}^{}\frac{dy}{1+\cos(4y)}$ 2 years ago 2 years ago
$1+\cos(4y)=2\cos^2(2y)$


text:
# [OS X TeX] Re: Issue with Multiple Bibliographies, MacOSX-TeX Digest #1826 - 08/19/06

Christopher Brislawn brislawn at lanl.gov
Mon Aug 21 11:52:47 EDT 2006

Begin forwarded message:
>
> Subject: Issue with Multiple Bibliographies
> From: "Simon Lisney" <simon at lisney.co.uk>
> Date: Sat, 19 Aug 2006 23:38:36 +0200
>
> I have recently installed MacTex and using TexShop in conjunction
> with Bibdesk for beginning to compile my thesis.
> So far I have been able to get it all working together and
> producing .tex documents with .bib files.  The problem comes when I
> try to create multiple bibliographies in one .tex file.
> <snip>
>
> I have read that there is an issue using Texshop with multibib but
> that it can be done. Being a beginner I am just not seeing the
> light.  I posted more detail on the Texshop forum which give more
> info here;
>
> http://www.apfelwiki.de/forum/viewtopic.php?
> t=1048&sid=10dd9b656cf5abd69572b2e95d3bb089
>
> If anyone can offer a way in which I can get Texshop to work with
> multiple bibs it would be great. Or offer another working solution

I have had good luck creating a resume' with multiple bibliography
sections (one for journal articles, one for conference papers, one
for book chapters, etc.) using the splitbib package:
http://tug.ctan.org/tex-archive/macros/latex/contrib/splitbib/

It works fine with BibTeX, and I have encountered no problems using
it with multiple .bib files under TeXShop 2.09/gwTeX.  The basic
approach of splitbib is to have the user define "category"
environments that list which bibliography a citation will appear in.
After that, you just cite references as usual in your document using
\cite{}, and splitbib handles the reorganization of citations into
multiple bibliography sections.  For instance, my resume' contains:

\begin{category}[J]{Journal Articles}
\SBentries{BrisWohl06b,BrisWohl06,...}
\end{category}

\begin{category}[B]{Book Chapters}
\SBentries{BCHQT03,ClodTheiBris03,...}
\end{category}

The first category results in a bibliography titled "Journal
Articles" with citation tags J1, J2,... for citations corresponding
to the citation labels BrisWohl06b, BrisWohl06,....  The second
category yields a second bibliography titled "Book Chapters" with
entries numbered B1, B2,... for the citations labelled BCHQT03,
ClodTheiBris03,..., and so on.

splitbib is a document-formatting package rather than a bibliographic
database tool, so it is indifferent to the method used to generate
references; it just rearranges them into the desired set of
bibliography sections.  It doesn't redefine the \cite command so it
is easy to add or remove splitbib formatting in existing documents or
change the categories in which the citations will appear.  It also
seems to work fine with various document styles.  The main cost is
the effort associated with manually listing your citations in the
desired categories.

Christopher M. Brislawn
Los Alamos National Laboratory
Computer & Computational Sciences
Group CCS-3, Mail Stop B265
Los Alamos, NM 87545-1663
brislawn at lanl.gov

------------------------- Info --------------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Archive: http://tug.org/pipermail/macostex-archives/


text:
# Linear Autocorrelation Function

These features capture properties of the linear autocorrelation function

## `CO_f1ecac` (`first1e_acf_tau)`

NB: The name `CO_f1ecac` derives from an earlier version of hctsa (the current version of hctsa names this feature as `first1e_acf_tau`).

### What it does

The `first1e_acf_tau` feature in catch22 computes the first 1/e crossing of the autocorrelation function of the time series. In hctsa, this can be computed as `CO_FirstCrossing(x_z,'ac',1/exp(1),'discrete')`.
This feature measures the first time lag at which the autocorrelation function drops below 1/e (= 0.3679).

### What it measures

`first1e_acf_tau` captures the approximate scale of autocorrelation in a time series. This can be thought of as the number of steps into the future at which a value of the time series at the current point and that future point remain substantially (>1/e) correlated. For a continuous-time system, this statistic is high when the sampling rate is high relative to the timescale of the dynamics.
• For uncorrelated noise, like the Poisson-distributed series shown below, the autocorrelation function drops to ~0 immediately, and we obtain the minimum value of this statistic: `first1e_acf_tau = 1`.
• For processes with a greater level of autocorrelation, the autocorrelation function decays more slowly, and we can obtain a larger value of this feature. Take this series simulated from a Chirikov map, which has `first1e_acf_tau = 6`
:
• We obtain even larger values for even more slowly varying time series, like this ODE, measured at a very high sampling rate, yielding `first1e_acf_tau = 17`
• Financial series (and many non-stationary stochastic processes) are highly autocorrelated, like this series for which `first1e_acf_tau = 176`

## `CO_FirstMin_ac` (`firstMin_acf)`

This feature is named `CO_FirstMin_ac` in catch22 and matches the hctsa feature named `firstMin_acf`.
Similar to the 1/e crossing feature above, `firstMin_acf` computes the first minimum of the autocorrelation function. It exhibits similar behavior.


text:
## 103.3 The lisse-étale and the flat-fppf sites

The section is the analogue of Cohomology of Stacks, Section 102.14 for derived categories.

Lemma 103.3.1. Let $\mathcal{X}$ be an algebraic stack. Notation as in Cohomology of Stacks, Lemmas 102.14.2 and 102.14.4.

1. The functor $g_! : \textit{Ab}(\mathcal{X}_{lisse,{\acute{e}tale}}) \to \textit{Ab}(\mathcal{X}_{\acute{e}tale})$ has a left derived functor

$Lg_! : D(\mathcal{X}_{lisse,{\acute{e}tale}}) \longrightarrow D(\mathcal{X}_{\acute{e}tale})$

which is left adjoint to $g^{-1}$ and such that $g^{-1}Lg_! = \text{id}$.

2. The functor $g_! : \textit{Mod}(\mathcal{X}_{lisse,{\acute{e}tale}}, \mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}}) \to \textit{Mod}(\mathcal{X}_{\acute{e}tale}, \mathcal{O}_{\mathcal{X}})$ has a left derived functor

$Lg_! : D(\mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}}) \longrightarrow D(\mathcal{X}_{\acute{e}tale}, \mathcal{O}_\mathcal {X})$

which is left adjoint to $g^*$ and such that $g^*Lg_! = \text{id}$.

3. The functor $g_! : \textit{Ab}(\mathcal{X}_{flat,fppf}) \to \textit{Ab}(\mathcal{X}_{fppf})$ has a left derived functor

$Lg_! : D(\mathcal{X}_{flat, fppf}) \longrightarrow D(\mathcal{X}_{fppf})$

which is left adjoint to $g^{-1}$ and such that $g^{-1}Lg_! = \text{id}$.

4. The functor $g_! : \textit{Mod}(\mathcal{X}_{flat,fppf}, \mathcal{O}_{\mathcal{X}_{flat,fppf}}) \to \textit{Mod}(\mathcal{X}_{fppf}, \mathcal{O}_{\mathcal{X}})$ has a left derived functor

$Lg_! : D(\mathcal{O}_{\mathcal{X}_{flat, fppf}}) \longrightarrow D(\mathcal{O}_\mathcal {X})$

which is left adjoint to $g^*$ and such that $g^*Lg_! = \text{id}$.

Warning: It is not clear (a priori) that $Lg_!$ on modules agrees with $Lg_!$ on abelian sheaves, see Cohomology on Sites, Remark 21.37.3.

Proof. The existence of the functor $Lg_!$ and adjointness to $g^*$ is Cohomology on Sites, Lemma 21.37.2. (For the case of abelian sheaves use the constant sheaf $\mathbf{Z}$ as the structure sheaves.) Moreover, it is computed on a complex $\mathcal{H}^\bullet$ by taking a suitable left resolution $\mathcal{K}^\bullet \to \mathcal{H}^\bullet$ and applying the functor $g_!$ to $\mathcal{K}^\bullet$. Since $g^{-1}g_!\mathcal{K}^\bullet = \mathcal{K}^\bullet$ by Cohomology of Stacks, Lemmas 102.14.4 and 102.14.2 we see that the final assertion holds in each case. $\square$

Lemma 103.3.2. With assumptions and notation as in Cohomology of Stacks, Lemma 102.15.1. We have

$g^{-1} \circ Rf_* = Rf'_* \circ (g')^{-1} \quad \text{and}\quad L(g')_! \circ (f')^{-1} = f^{-1} \circ Lg_!$

on unbounded derived categories (both for the case of modules and for the case of abelian sheaves).

Proof. Let $\tau = {\acute{e}tale}$ (resp. $\tau = fppf$). Let $\mathcal{F}$ be an abelian sheaf on $\mathcal{X}_\tau$. By Cohomology of Stacks, Lemma 102.15.3 the canonical (base change) map

$g^{-1}Rf_*\mathcal{F} \longrightarrow Rf'_*(g')^{-1}\mathcal{F}$

is an isomorphism. The rest of the proof is formal. Since cohomology of abelian groups and sheaves of modules agree we also conclude that $g^{-1} Rf_*\mathcal{F} = Rf'_* (g')^{-1}\mathcal{F}$ when $\mathcal{F}$ is a sheaf of modules on $\mathcal{X}_\tau$.

Next we show that for $\mathcal{G}$ (either sheaf of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{flat,fppf}$) the canonical map

$L(g')_!(f')^{-1}\mathcal{G} \to f^{-1}Lg_!\mathcal{G}$

is an isomorphism. To see this it is enough to prove for any injective sheaf $\mathcal{I}$ on $\mathcal{X}_\tau$ the induced map

$\mathop{\mathrm{Hom}}\nolimits (L(g')_!(f')^{-1}\mathcal{G}, \mathcal{I}[n]) \leftarrow \mathop{\mathrm{Hom}}\nolimits (f^{-1}Lg_!\mathcal{G}, \mathcal{I}[n])$

is an isomorphism for all $n \in \mathbf{Z}$. (Hom's taken in suitable derived categories.) By the adjointness of $f^{-1}$ and $Rf_*$, the adjointness of $Lg_!$ and $g^{-1}$, and their “primed” versions this follows from the isomorphism $g^{-1} Rf_*\mathcal{I} \to Rf'_* (g')^{-1}\mathcal{I}$ proved above.

In the case of a bounded complex $\mathcal{G}^\bullet$ (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$) the canonical map

103.3.2.1
$$\label{stacks-perfect-equation-to-show} L(g')_!(f')^{-1}\mathcal{G}^\bullet \to f^{-1}Lg_!\mathcal{G}^\bullet$$

is an isomorphism as follows from the case of a sheaf by the usual arguments involving truncations and the fact that the functors $L(g')_!(f')^{-1}$ and $f^{-1}Lg_!$ are exact functors of triangulated categories.

Suppose that $\mathcal{G}^\bullet$ is a bounded above complex (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$). The canonical map (103.3.2.1) is an isomorphism because we can use the stupid truncations $\sigma _{\geq -n}$ (see Homology, Section 12.15) to write $\mathcal{G}^\bullet$ as a colimit $\mathcal{G}^\bullet = \mathop{\mathrm{colim}}\nolimits \mathcal{G}_ n^\bullet$ of bounded complexes. This gives a distinguished triangle

$\bigoplus \nolimits _{n \geq 1} \mathcal{G}_ n^\bullet \to \bigoplus \nolimits _{n \geq 1} \mathcal{G}_ n^\bullet \to \mathcal{G}^\bullet \to \ldots$

and each of the functors $L(g')_!$, $(f')^{-1}$, $f^{-1}$, $Lg_!$ commutes with direct sums (of complexes).

If $\mathcal{G}^\bullet$ is an arbitrary complex (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$) then we use the canonical truncations $\tau _{\leq n}$ (see Homology, Section 12.15) to write $\mathcal{G}^\bullet$ as a colimit of bounded above complexes and we repeat the argument of the paragraph above.

Finally, by the adjointness of $f^{-1}$ and $Rf_*$, the adjointness of $Lg_!$ and $g^{-1}$, and their “primed” versions we conclude that the first identity of the lemma follows from the second in full generality. $\square$

Lemma 103.3.3. Let $\mathcal{X}$ be an algebraic stack. Notation as in Cohomology of Stacks, Lemma 102.14.2.

1. Let $\mathcal{H}$ be a quasi-coherent $\mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}}$-module on the lisse-étale site of $\mathcal{X}$. For all $p \in \mathbf{Z}$ the sheaf $H^ p(Lg_!\mathcal{H})$ is a locally quasi-coherent module with the flat base change property on $\mathcal{X}$.

2. Let $\mathcal{H}$ be a quasi-coherent $\mathcal{O}_{\mathcal{X}_{flat,fppf}}$-module on the flat-fppf site of $\mathcal{X}$. For all $p \in \mathbf{Z}$ the sheaf $H^ p(Lg_!\mathcal{H})$ is a locally quasi-coherent module with the flat base change property on $\mathcal{X}$.

Proof. Pick a scheme $U$ and a surjective smooth morphism $x : U \to \mathcal{X}$. By Modules on Sites, Definition 18.23.1 there exists an étale (resp. fppf) covering $\{ U_ i \to U\} _{i \in I}$ such that each pullback $f_ i^{-1}\mathcal{H}$ has a global presentation (see Modules on Sites, Definition 18.17.1). Here $f_ i : U_ i \to \mathcal{X}$ is the composition $U_ i \to U \to \mathcal{X}$ which is a morphism of algebraic stacks. (Recall that the pullback “is” the restriction to $\mathcal{X}/f_ i$, see Sheaves on Stacks, Definition 95.9.2 and the discussion following.) After refining the covering we may assume each $U_ i$ is an affine scheme. Since each $f_ i$ is smooth (resp. flat) by Lemma 103.3.2 we see that $f_ i^{-1}Lg_!\mathcal{H} = Lg_{i, !}(f'_ i)^{-1}\mathcal{H}$. Using Cohomology of Stacks, Lemma 102.8.2 we reduce the statement of the lemma to the case where $\mathcal{H}$ has a global presentation and where $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ for some affine scheme $X = \mathop{\mathrm{Spec}}(A)$.

Say our presentation looks like

$\bigoplus \nolimits _{j \in J} \mathcal{O} \longrightarrow \bigoplus \nolimits _{i \in I} \mathcal{O} \longrightarrow \mathcal{H} \longrightarrow 0$

where $\mathcal{O} = \mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}}$ (resp. $\mathcal{O} = \mathcal{O}_{\mathcal{X}_{flat,fppf}}$). Note that the site $\mathcal{X}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{X}_{flat,fppf}$) has a final object, namely $X/X$ which is quasi-compact (see Cohomology on Sites, Section 21.16). Hence we have

$\Gamma (\bigoplus \nolimits _{i \in I} \mathcal{O}) = \bigoplus \nolimits _{i \in I} A$

by Sites, Lemma 7.17.7. Hence the map in the presentation corresponds to a similar presentation

$\bigoplus \nolimits _{j \in J} A \longrightarrow \bigoplus \nolimits _{i \in I} A \longrightarrow M \longrightarrow 0$

of an $A$-module $M$. Moreover, $\mathcal{H}$ is equal to the restriction to the lisse-étale (resp. flat-fppf) site of the quasi-coherent sheaf $M^ a$ associated to $M$. Choose a resolution

$\ldots \to F_2 \to F_1 \to F_0 \to M \to 0$

by free $A$-modules. The complex

$\ldots \mathcal{O} \otimes _ A F_2 \to \mathcal{O} \otimes _ A F_1 \to \mathcal{O} \otimes _ A F_0 \to \mathcal{H} \to 0$

is a resolution of $\mathcal{H}$ by free $\mathcal{O}$-modules because for each object $U/X$ of $\mathcal{X}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{X}_{flat,fppf}$) the structure morphism $U \to X$ is flat. Hence by construction the value of $Lg_!\mathcal{H}$ is

$\ldots \to \mathcal{O}_\mathcal {X} \otimes _ A F_2 \to \mathcal{O}_\mathcal {X} \otimes _ A F_1 \to \mathcal{O}_\mathcal {X} \otimes _ A F_0 \to 0 \to \ldots$

Since this is a complex of quasi-coherent modules on $\mathcal{X}_{\acute{e}tale}$ (resp. $\mathcal{X}_{fppf}$) it follows from Cohomology of Stacks, Proposition 102.8.1 that $H^ p(Lg_!\mathcal{H})$ is quasi-coherent. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
AIMS Mathematics, 2020, 5(3): 2758-2779. doi: 10.3934/math.2020178.

Research article Special Issues

Export file:

Format

• RIS(for EndNote,Reference Manager,ProCite)
• BibTex
• Text

Content

• Citation Only
• Citation and Abstract

Parameter estimation and fractional derivatives of dengue transmission model

1 Department of Mathematics, Faculty of Science and Technology, Universitas Airlangga, Surabaya 60115, Indonesia
2 Faculty of Natural and Agricultural Sciences, University of the Free state, South Africa

## Abstract    Full Text(HTML)    Figure/Table    Related pages

In this paper, we propose a parameter estimation of dengue fever transmission model using a particle swarm optimization method. This method is applied to estimate the parameters of the host-vector and SIR type dengue transmission models by using cumulative data of dengue patient in East Java province, Indonesia. Based on the parameter values, the basic reproduction number of both models are greater than one and obtained their value for SIR is $\mathcal{R}_0=1.4159$ and for vector host is $\mathcal{R}_0=1.1474$. We then formulate the models in fractional Atangana-Baleanu derivative that possess the property of nonlocal and nonsingular kernel that has been remained effective to many real-life problems. A numerical procedure for the solution of the model SIR model is shown. Some specific numerical values are considered to obtain the graphical results for both the SIR and Vector Host model. We show that the model vector host provide good results for data fitting than that of the SIR model.
Figure/Table
Supplementary
Article Metrics

Citation: Windarto, Muhammad Altaf Khan, Fatmawati. Parameter estimation and fractional derivatives of dengue transmission model. AIMS Mathematics, 2020, 5(3): 2758-2779. doi: 10.3934/math.2020178

References

• 1. World Health Organization, Fact sheet on the Dengue and severe dengue, WHO, 2017. Available from: http://www.who.int/mediacentre/factsheets/fs117/en/.
• 2. World Health Organization, Global strategy for dengue prevention and control 2012-2020, WHO, 2012. Available from: http://www.who.int/denguecontrol/9789241504034/en/.
• 3. Health Office (Dinas Kesehatan) of the East Java, Dinas Kesehatan Provinsi Jawa Timur, Surabaya, Indonesia, 2009.
• 4. Ministry of Health Republic of Indonesia, Kementerian Kesehatan Republik Indonesia Jakarta, 2014.
• 5. D. Aldila, T. Gotz, E. Soewono, An optimal control problem arising from a dengue disease transmission model, Math. Biosci., 242 (2013), 9-16.
• 6. H. Tasman, A. K. Supriatna, N. Nuraini, et al. A dengue vaccination model for immigrants in a two-age-class population, Int. J. Math., 2012 (2012), 1-15.
• 7. J. P. Chavez, T. Gotz, S. Siegmund, et al. An SIR-Dengue transmission model with seasonal effects and impulsive control, Math. Biosci., 289 (2017), 29-39.
• 8. A. Pandey, A. Mubayi, J. Medlock, Comparing vector-host and SIR models for dengue transmission, Math. Biosci., 246 (2013), 252-259.
• 9. T. Gotz, N. Altmeier, W. Bock, et al. Modeling dengue data from Semarang, Indonesia, Ecol. Complex., 30 (2017), 57-62.
• 10. F. B. Agusto, M. A. Khan, Optimal control strategies for dengue transmission in pakistan, Math. Biosci., 305 (2018), 102-121.
• 11. N. Tutkun, Parameter estimation in mathematical models using the real coded genetic algorithms, Expert Syst. Appl., 36 (2009), 3342-3345.
• 12. R. L. Haupt, S. E. Haupt, Practical Genetic Algorithms, Second Edition, John Wiley & Sons, 2004.
• 13. W. B. Roush, S. L. Branton, A Comparison of fitting growth models with a genetic algorithm and nonlinear regression, Poultry Sci., 84 (2005), 494-502.
• 14. W. S. W. Indratno, N. Nuraini, E. Soewono, A comparison of binary and continuous genetic algorithm in parameter estimation of a logistic growth model, American Institute of Physics Conference Series, 1587 (2014), 139-142.
• 15. Windarto, An Implementation of continuous genetic algorithm in parameter estimation of predator-prey model, AIP Conference Proceedings, 2016.
• 16. D. Akman, O. Akman, E. Schaefer, Parameter Estimation in Ordinary Differential Equations Modeling via Particle Swarm Optimization, J. Appl. Math., 2018 (2018), 1-9.
• 17. Windarto, Eridani, U. D. Purwati, A comparison of continuous genetic algorithm and particle swarm optimization in parameter estimation of Gompertz growth model, AIP Conference Proceedings, 2084 (2019), 020017.
• 18. R. Eberhart, J. Kennedy, A new optimizer using particle swarm theory, Proceedings of the Sixth International Symposium on Micro Machine and Human Science, 1995.
• 19. K. Diethelm, A fractional calculus based model for the simulation of an outbreak of dengue fever, Nonlinear Dyn., 71 (2013), 613-619.
• 20. T. Sardar, S. Rana, J. Chattopadhyay, A mathematical model of dengue transmission with memory, Commun. Nonlinear Sci., 22 (2015), 511-525.
• 21. M. A. Khan, S. Ullah, M. Farooq, A new fractional model for tuberculosis with relapse via Atangana-Baleanu derivative, Chaos, Solitons & Fractals, 116 (2018), 227-238.
• 22. S. Ullah, M. A. Khan, M. Farooq, Modeling and analysis of the fractional HBV model with Atangana-Baleanu derivative, The European Physical Journal Plus, 133 (2018), 313.
• 23. E. O. Alzahrani, M. A. Khan, Modeling the dynamics of Hepatitis E with optimal control, Chaos, Solitons & Fractals, 116 (2018), 287-301.
• 24. Fatmawati, E. M. Shaiful, M. I. Utoyo, A fractional order model for HIV dynamics in a two-sex population, Int. J. Math., 2018 (2018), 1-11.
• 25. Fatmawati, M. A. Khan, M. Azizah, et al. A fractional model for the dynamics of competition between commercial and rural banks in Indonesia, Chaos, Solitons & Fractals, 122 (2019), 32-46.
• 26. A. Atangana, D. Baleanu, New fractional derivatives with nonlocal and non-singular kernel: theory and application to heat transfer model, Therm. Sci., 20 (2016), 763-769.
• 27. S. Qureshi, A. Atangana, Mathematical analysis of dengue fever outbreak by novel fractional operators with field data, Physica A, 526 (2019), 121127.
• 28. O. Diekmann, J. A. P. Heesterbeek, J. A. J. Metz, On the definition and the computation of the basic reproduction ratio R0 in models for infectious diseases in heterogenous populations, J. Math. Biol., 28 (1990), 362-382.
• 29. O. Diekmann, J. A. P. Heesterbeek, Mathematical Epidemiology of Infectious Diseases, Model Building, Analysis and Interpretation, John Wiley & Son, 2000.
• 30. P. van den Driessche, J. Watmough, Reproduction numbers and sub-threshold endemic equilibria for compartmental models of disease transmition, Math. Biosci., 180 (2002), 29-48.
• 31. Wikipedia contributors: East Java, Wikipedia. Available from: https://en.wikipedia.org/wiki/East-Java.
• 32. M. Toufik, A. Atangana, New numerical approximation of fractional derivative with non-local and non-singular kernel: application to chaotic models, Eur. Phys. J. Plus, 132 (2017), 444.

• 1. Yanmei Wang, Guirong Liu, Stability of a Nonlinear Stochastic Epidemic Model with Transfer from Infectious to Susceptible, Complexity, 2020, 2020, 1, 10.1155/2020/9614670
• 2. Afeez Abidemi, Nur Arina Bazilah Aziz, Optimal control strategies for dengue fever spread in Johor, Malaysia, Computer Methods and Programs in Biomedicine, 2020, 105585, 10.1016/j.cmpb.2020.105585
• 3. Kankan Sarkar, Subhas Khajanchi, Juan J. Nieto, Modeling and forecasting the COVID-19 pandemic in India, Chaos, Solitons & Fractals, 2020, 139, 110049, 10.1016/j.chaos.2020.110049
• 4. Prasantha Bharathi Dhandapani, Dumitru Baleanu, Jayakumar Thippan, Vinoth Sivakumar, On stiff, fuzzy IRD-14 day average transmission model of COVID-19 pandemic disease, AIMS Bioengineering, 2020, 7, 4, 208, 10.3934/bioeng.2020018