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Question 5.6

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TZ
leumasicOfficial

3 months ago

i)

Let

δ1=sin2(9ϵ24)+ϵ2andδ2=ϵ24.\delta_1 = \sin^2\left(\frac{9\epsilon^2}{4}\right)+\frac{\epsilon}{2} \qquad\text{and}\qquad \delta_2 = \frac{\epsilon^2}{4}.

Then, δ=min(δ1,δ2)\delta = \min(\delta_1,\delta_2).

ii)

Let

δ1=min(sin2(14)+1,  sin2(9ϵ2100)+ϵ10)andδ2=ϵ236.\delta_1 = \min\left(\sin^2\left(\frac{1}{4}\right)+1,\; \sin^2\left(\frac{9\epsilon^2}{100}\right)+\frac{\epsilon}{10}\right) \qquad\text{and}\qquad \delta_2 = \frac{\epsilon^2}{36}.

Then, δ=min(δ1,δ2)\delta = \min(\delta_1,\delta_2).

iii)

δ=min(4,64ϵ2).\delta = \min(4, 64\epsilon^2).

iv)

Let

δ1=min(sin2(19)+1,  sin2(9ϵ225)+4ϵ10)andδ2=min(4,16ϵ2).\delta_1 = \min\left(\sin^2\left(\frac{1}{9}\right)+1,\; \sin^2\left(\frac{9\epsilon^2}{25}\right)+\frac{4\epsilon}{10}\right) \qquad\text{and}\qquad \delta_2 = \min(4, 16\epsilon^2).

Then, δ=min(δ1,δ2)\delta = \min(\delta_1,\delta_2).

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Q 5.6

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Q 5.6