Wikipedia:Reference desk/Archives/Mathematics/2008 August 26

= August 26 =

Question 1
Now i know how to do this one. Some of the underlying fundamentals i do not understand. (I've just been guessing the pronumeral and checking if it gives a set of numbers that have the same difference between them & substituting them in.. a better method would be appreciated)

Heres an example of one, that i know how to solve:

Find x, the first term, the common difference and the 4th term: {x + 5, 2x + 10, 3x + 15, ...}

I messed around with some numbers and i found x was any real number, the common difference is x+5 and the 4th term is 4x + 20.

This is another one but it gives not so nice fractional values for the pro-numerals in the answer section.

{x + 2, 4x + 4, 10x + 1, ...}

How would one determine the above required values in this case (Well mainly X, the rest would just come)?


 * If there's a common difference, that means (10x + 1) - (4x + 4) = (4x + 4) - (x + 2), which shouldn't be too hard to solve. -- Jao (talk) 14:00, 26 August 2008 (UTC)

Question 2
Find the number of terms in each of the following sequences:

{-0.8, -0.9, -1,..., -3.9}

According to the written material the formula is: Tn = a + (n-1)d

Tn = Last term in the sequence (-3.9)

a = First Term in the sequence (-0.8)

d = Difference Between Terms (-0.1)

So:

-3.9 = -0.8 + (n-1)-0.1

-3.1 =(n - 1)-0.1

31= n -1

n = 32

Now that the example i done is over, time for the actual question im having trouble with

{Log 2, Log 8, Log 32, ..., 37 Log 2}

Tn= 37 Log 2

A = Log 2

D = (Previous term multiplied by 4 perhaps?)

37 Log 2 = Log 2 + (n - 1)d

And im stuck there.. also im a bit confused to what the 37 infront of the log means.

I only have advanced math once a week and its been a while since i did logarithms

Cheers, Kingpomba (talk) 13:30, 26 August 2008 (UTC).


 * D = Log 8 - Log 2 = 3 Log 2 - Log 2 = 2 Log 2. And "37 Log 2" is simply a multiplication of 37 and Log 2. -- Jao (talk) 14:04, 26 August 2008 (UTC)


 * The sequence seems to go like:

{log(2^1), log (2^3), log(2^5),....., log(2^37)}

so there is a one-to-one correspondence between the sequence and the odd numbers from 1 to 37, i.e the number of terms in the sequence equals the number of odd integers from 1 to 37. There are 19 = (37 + 1)/2 such odd integers so there are 19 terms in the sequence.

Note: log (2^37) = 37*log(2) ; this is a property of logs.

Hope this helps.

Topology Expert (talk) 06:31, 27 August 2008 (UTC)

Calculus
Why is calculus sometimes referred to as "the calculus"? What exactly does that mean? Is there any distinction between "calculus" and "the calculus"? Is one term correct / preferred? Thanks. (Joseph A. Spadaro (talk) 15:10, 26 August 2008 (UTC))


 * There are other meanings of the word such as lambda calculus or calculus of variations. In modern usage, calculus is the same as "the" calculus, but for emphasis, clarity, and to be old-school, people like to say "the calculus". —Ben FrantzDale (talk) 16:03, 26 August 2008 (UTC)


 * I've never heard of calculus referred to as 'the calculus' on its own, only in terms like 'the calculus of variations' - and even then not often. Where have you seen it referred to this way thanks? It sounds to me like the sort of thing somebody from a couple of hundred years ago might have said. Dmcq (talk) 17:50, 26 August 2008 (UTC)
 * I've had a better look around and it seems I've just not noticed 'The calculus' being used this way. Interesting question, why have people chosen one or the other. I'd guess 'The Calculus' is older and is like 'The Principia' or 'The Method' and only later did i become commonplace.Dmcq (talk) 18:37, 26 August 2008 (UTC)


 * I use "calculus" to generically mean "a body of techniques for solving a particular class of problems", and "the calculus" to specifically mean "a body of techniques for solving differential and integral problems", i.e., calculus. Although I do frequently use "calculus" for the specific meaning, as well, if context makes the meaning clear.  Eric.  90.184.71.189 (talk) 10:29, 27 August 2008 (UTC)

Kernel density estimation question
A question has been sitting on the talk page of Kernel density estimation regarding the choice of kernel. Could someone take a crack at answering it? —Ben FrantzDale (talk) 16:05, 26 August 2008 (UTC)

Limits
I am having trouble determining this limit because it involves both x and y. I know that the answer is $$e^3$$. Could someone help by explaining it please? It is part of a wider question but I don't think the rest is relevant.


 * $$ \lim_{x \to -\infty} y e^{-x} $$

Thanks 92.5.221.82 (talk) 19:06, 26 August 2008 (UTC)


 * Is x going to infinity or negative infinity? What is y doing?  Wanderer57 (talk) 19:21, 26 August 2008 (UTC)

Assuming y is independent of x, we have
 * $$ \lim_{x \to -\infty} y e^{-x} =y \lim_{x \to -\infty} e^{-x} =  y \lim_{x \to \infty} e^{x} =

\begin{cases} -\infty & y <0 \\ 0 & y = 0 \\ \infty & y >0 \end{cases} $$ If y is dependent on x, then you'd have to know how in order to compute the limit. &mdash; Carl (CBM · talk) 19:28, 26 August 2008 (UTC)


 * Carl, the middle part of your answer is incorrect — it's not true, that $$y \lim_{x \to \infty} e^x = 0$$ for y being zero. In that case you have $$y \lim_{x \to \infty} e^x = 0\cdot\infty$$ which is undefined. To make the solution correct split it into two parts, for y being zero and non-zero, respectively. Then you have $$y=0 \ \Rightarrow\ \lim_{x \to \infty} ye^x = \lim_{x \to \infty} 0 = 0$$ and $$y\ne 0 \ \Rightarrow\  \lim_{x \to \infty} ye^x = \begin{cases}

-|y| \cdot \infty & \mathrm{for\ }y<0 \\ -\infty & \mathrm{for\ }y<0 \\ \infty & \mathrm{for\ }y>0 \end{cases}$$ CiaPan (talk) 13:47, 27 August 2008 (UTC)
 * y| \cdot \infty & \mathrm{for\ }y>0 \end{cases} = \begin{cases}


 * Of course you'd get an indeterminate form $$0 \cdot \infty$$; but then examination of the initial limit shows that in this particular situation the the limit value is 0. I was somewhat terse. &mdash; Carl (CBM · talk) 14:17, 27 August 2008 (UTC)


 * Reducing formula to an indeterminate form, then proceedeing through the initial formula to assign a specific value to the $$0 \cdot \infty$$ seems 'simpler than possible' to me.
 * I thought that in this case there's no 'then' and no 'particular situation'. When an expression evaluates to an undefined symbol, one needs to get back, exclude the particular case from the general solution and solve it separately. That's what my teachers used to say, AFAIR (however it was long, long ago....). --CiaPan (talk) 12:13, 28 August 2008 (UTC)
 * Excluding that particular case and solving it separately is exactly what I mean by "examination of the initial limit". You're right that if I had actually written a proof (in English) rather than just written an equation (which is not a proof), the proof would have specifically dealt with the case y = 0. &mdash; Carl (CBM · talk) 15:26, 28 August 2008 (UTC)