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This is probably me being a bit stupid and not noticing why, but I'm really stuck on this partition question from analysis... I understand the technique it's just the values that are correct - I'm not sure why?

f(x) : [-1,1] \rightarrow \mathbb{R}

defined by

Unparseable latex formula:

f(x) = \begin{Bmatrix} 2, x \in [-1,0)\\-1,]

P_n = {-1, \frac{-1}{n} , \frac{1}{n} ,1}

When working out the

m_1 and M_1

values, these are the answers...

m_1 = inf ({f(x) :x \in [-1,\frac{-1}{n}]}) = 2[br]m_2 = inf ({f(x) : x \in [\frac{-1}{n},\frac{1}{n}]}) = -1 [br]m_3 = inf ({f(x) : x \in [\frac{1}{n},1]}) = 1

M_1 = sup ({f(x) :x \in [-1,\frac{-1}{n}]}) = 2[br]M_2 = sup ({f(x) : x \in [\frac{-1}{n},\frac{1}{n}]}) = 2[br]M_3 = sup ({f(x) : x \in [\frac{1}{n},1]}) = 1

It's just these values I can't get my head around, I don't understand for example why

m_2

is -1 instead of 2? I was thinking the infimum of

[\frac{-1}{n},1]

would be -1?

Thanks for any help!

(Apologies for the terrible latex attempt... :rolleyes:

)

(edited 9 years ago)

Reply 1

ghostwalker

Original post by Blue7195

It's just these values I can't get my head around, I don't understand for example why

m_2

is -1 instead of 2?

The subinterval for

m_2

contains the point x=0, for which f(x)=-1

Can't be sure of the answer for

m_3

as the details of f(x) are missing for the final subinterval.

(edited 9 years ago)

Reply 2

Blue7195

Original post by ghostwalker

The subinterval for

m_2

contains the point x=0, for which f(x)=-1

Can't be sure of the answer for

m_3

as the details of f(x) are missing for the final subinterval.

= 1, x \in (0,1]

sorry about that!
And thanks for your help

Reply 3

ghostwalker

Original post by Blue7195

= 1, x \in (0,1]

OK, then for

m_3

, f(x) is 1 on the whole subinterval [1/n,1], so the inf is also 1.

All clear now?

Reply 4

Blue7195

Original post by ghostwalker

OK, then for

m_3

, f(x) is 1 on the whole subinterval [1/n,1], so the inf is also 1.

All clear now?

Yeah thank you! :smile:

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