-- Sep 13 In-Class Exercise
`m=max(\frac{144ln\delta}{2},m_i)`
`m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}`
Given,
`\delta=\frac{1}{1000}`
`\epsilon=\frac{1}{n+1}`
Thus,
`m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)`
`m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }`
Since, ln(1/1000) / 2 < 0,
`m = (144*(n+1)^4) / 25*k^2`
Thus, no. of samples
`= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)`
`= (288/ 25) * ( (n+1)^5 / k^2)`
k = 1 / (c * 2) = 1 / (3 * 2) = 1 / 6
Thus, no. of samples:
`= (288 * 36/ 25) * ( (n+1)^5 )`
(
Edited: 2017-10-02)
@BT@m=max(\frac{144ln\delta}{2},m_i)@BT@
@BT@m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}@BT@
Given,
@BT@\delta=\frac{1}{1000}@BT@
@BT@\epsilon=\frac{1}{n+1}@BT@
Thus,
@BT@m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)@BT@
@BT@m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }@BT@
Since, ln(1/1000) / 2 < 0,
@BT@m = (144*(n+1)^4) / 25*k^2@BT@
Thus, no. of samples
@BT@= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)@BT@
@BT@= (288/ 25) * ( (n+1)^5 / k^2)@BT@
k = 1 / (c * 2) = 1 / (3 * 2) = 1 / 6
Thus, no. of samples:
@BT@= (288 * 36/ 25) * ( (n+1)^5 )@BT@