-- Sep 13 In-Class Exercise
`m=max(\frac{144ln\delta}{2},m_i)`
`m_1= \frac{144(n+1)^2} {25(k\epsilon)^2}`
provided,
`\epsilon = \frac{1} {n+1}`
and,
`\delta = \frac{1} {1000}`
since,
inserting delta in m, 1st component would be negative and m1 would result in maximum
m = `m_1`
`k = frac{1}{2c}`
since,
c=3
`k = frac{1}{6}`
substituting value of k and `epsilon` in equation gives
`m_1 = \frac{144((6)^2)(n+1)^4}{25}`
(
Edited: 2017-09-13)
@BT@m=max(\frac{144ln\delta}{2},m_i)@BT@
@BT@m_1= \frac{144(n+1)^2} {25(k\epsilon)^2}@BT@
provided,
@BT@\epsilon = \frac{1} {n+1}@BT@
and,
@BT@\delta = \frac{1} {1000}@BT@
since,
inserting delta in m, 1st component would be negative and m1 would result in maximum<br/>
m = @BT@m_1@BT@<br/>
@BT@k = frac{1}{2c}@BT@<br/>
since, <br/>
c=3<br/>
@BT@k = frac{1}{6}@BT@<br/>
substituting value of k and @BT@epsilon@BT@ in equation gives<br/>
@BT@m_1 = \frac{144((6)^2)(n+1)^4}{25}@BT@