TT's Maths Thread

[TrueTears]

oooo nice man

so...

?

[/0]

lol yep

 

[humph]

tpa gki pussg??

[TrueTears]

That is so annoying, how to get around this...?

nvm I got it  :tickedoff:

[TrueTears]

When using logarithmic differentiation when do we use modulus?

Eg,



Now do we need modulus around the since it is negative for some ? As in:



What about when we simplify and get:



Originally is positive for all but now once you take out the , is negative for some , so do we need to do:



So yeah just confused when we need to put modulus...

[/0]

If you take the logs of both sides that's like where , so the domain of will be restricted to where .

However, you can apply the modulus first:





The derivative of is the same as the derivative of or

[TrueTears]

If you take the logs of both sides that's like where , so the domain of will be restricted to where .

However, you can apply the modulus first:





The derivative of is the same as the derivative of or

Thanks for that

y can't equal 0 right?

Since log(0) is undefined?

So shudn't it be y>0?

Also putting in modulus doesnt affect the derivative anyways so we can just omit it most of the time yeah?

[/0]

Yeah

[TrueTears]

Yeah

Oh okay, then what the hell is the point of worrying about if f(x)<0 anyway lulz, it's not like it's gonna affect the derivative haha

[TrueTears]

For the extreme value theorem, what if the function is a horizontal line in the interval .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

[Damo17]

For the extreme value theorem, what if the function is a horizontal line in the interval .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.

[TrueTears]

For the extreme value theorem, what if the function is a horizontal line in the interval .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.

mmm yeah thanks heaps , but isn't that a bit weird since there is no 'extreme' value (if you get what I mean)

Since here absolute min = absolute max, so there shouldn't be any extreme value, it just doesn't make any intuitive sense hahaa

[Damo17]

For the extreme value theorem, what if the function is a horizontal line in the interval .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.

mmm yeah thanks heaps , but isn't that a bit weird since there is no 'extreme' value (if you get what I mean)

Since here absolute min = absolute max, so there shouldn't be any extreme value, it just doesn't make any intuitive sense hahaa

It does seem weird to me, but it passes the requirements for the 'Extreme Value Theorem (as far as I can see), so that is the conclusion I draw from that. But others that know more than me will give a better answer. 

[humph]

Extreme value just means that there exists some such that for all (or something along those lines). There's no claim of uniqueness of an extreme value (hence the possible equality in the lesser-than-or-equal-to sign).
Often in complex analysis, you'll be given some complex-differentiable function and told behaviour of where it attains its maxima and minima, and you can pretty much determine the function from that (or at least very useful global behaviour). But that's because complex-differentiable functions behave incredibly nicely in general (whereas real-differentiable functions can be extremely poorly behaved).

[TrueTears]

Thanks ^^

Prove

I'm thinking of using MVT... but how though?