Logarithm notation

theclam

I'm doing some homework and the professor gave me this:

(Y^2*((log^5)Y))

I can't understand the notation.

I can see a bunch of different possibilities:

(log^5)Y could be:

log(log(log(log(log Y))))
log(Y^5)
(log Y)^5
log5 Y

Y^2*stuff could be:

(Y^2)*stuff
Y^(2*stuff)

For reference, the original question is:

What is the Big-Oh of K(Y) if
K(Y) = 9*K(Y/3)+(Y^2*((log^5)Y))

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clsCorwin

Is the 5 subscript or superscript? If its subscript, then its log base 5. If its superscript, its (log Y)^5, just like how they do with trigonometric functions.

Clipse

mcdermott wrote: »

The most likely for (log^5) is a log base 5, even though that seems pretty absurdly uncommon (I've mostly dealt with log base e, log base 10, or log base 2).

Y^2*stuff is almost certainly (Y^2)*stuff if he's following any sort of notation standard, because you always do the exponential before the multiplication.

I would guess (log^5)(Y) = (log(Y))^5 actually. Look, for instance, at equation 13 of this. Obviously it's cos, not log, but I see that notation used for exponents of the log function regularly.

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theclam

I copy/pasted it, so it's given exactly as shown.

I think I'm going to go with Y^(2*((log Y)^5)), since it makes the most sense in terms of figuring out an answer. I'm going to frame my answer in terms of (log^5)Y though, to see if I can get some partial credit for being ambiguous, rather than outright wrong.

Thanks for the help, I guess this can be locked.

Teslan26

The Big-oh? what in the hell is that?

K(Y) = 9*K(Y/3)+(Y^2*((log^5)Y))

I like how the function(?) has itself in its definition too, unless K is a constant in there, or are you trying to isolate K?

Odd, very very odd format.

HalberdBlue

Big-oh is the growth rate of the function.

cyphr

It's log(y) to the fifth power. Your question asks about big-O notation, which means this is for a CS class. The base of the logarithm is irrelevant in complexity theory because you can convert to any other base via some constant factor.

So if my complexity theory chops are still any good, I'm going to say that K(Y) is O(Y^2 * logY ^5 ).

(see here for more info: indefinite logarithms)

Teslan26

Ah, so Big-oh as in the order of magnitude sort of thing.

Papillon

theclam wrote: »

I copy/pasted it, so it's given exactly as shown.

I think I'm going to go with Y^(2*((log Y)^5)), since it makes the most sense in terms of figuring out an answer. I'm going to frame my answer in terms of (log^5)Y though, to see if I can get some partial credit for being ambiguous, rather than outright wrong.

Thanks for the help, I guess this can be locked.

It's almost certainly (Y^2)*((log Y)^5) instead of Y^(2*((log Y)^5)) since exponentiation is higher in the order of operations than multiplication.

Starcross

Email the professor about this if you aren't completely sure.