Wikipedia:Reference desk/Archives/Mathematics/2008 April 17

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April 17

Calculus

What is the integral of a partial derivative? (The integral of the full derivative of $f(x)$ is $f(x)$.) Also, how would this be written? $\textstyle {\int {\frac {\partial }{\partial {y}}}f(x)\ \mathrm {d} x}$? Also, can a partial integral be taken by replacing $\textstyle \operatorname {d} x\,\!$ with $\textstyle \partial {x}$? Thanks, Zrs 12 (talk) 02:15, 17 April 2008 (UTC)

In a general sense, all integrals are "partial integrals", in the sense that you hold one variable constant when integrating with respect to another. So $\int {f_{x}(x,y)\ \mathrm {d} x}=f(x,y)+C$. As far as I know there is no notation that replaces the total dx with a partial symbol in an integral. Confusing Manifestation(Say hi!) 06:56, 17 April 2008 (UTC)

The antiderivative on the other hand would be the integral of $f_{x}(x,y)$ plus some function purely in terms of y. -mattbuck (Talk) 11:21, 17 April 2008 (UTC)

Ah, of course. Very important when solving differential equations, particularly separable ones. Confusing Manifestation(Say hi!) 05:17, 18 April 2008 (UTC)

Applications of Cauchy's Integral Theorem and Formula

HOW TO SOLVE THIS PROBLEM?

INTEGRATE (Z²-Z+1)/Z-1 WITH RESPECT TO Z OVER THE UNIT CIRCLE —Preceding unsigned comment added by 59.93.24.178 (talk) 10:38, 17 April 2008 (UTC)

Hello, this question looks like homework, and as a rule the reference desk will not answer homework questions unless you have actually tried to answer them yourself. I suggest looking at the page Cauchy's integral formula. Also, please turn off your Caps lock button, as typing in capitals is generally construed as shouting. -mattbuck (Talk) 11:30, 17 April 2008 (UTC)

Assuming that your function is actually (z²-z+1)/(z-1), then you have a bit of a conundrum, as the pole at z=1 is actually on the unit circle. Now you might "tweak" the path to "wiggle" around the pole. But then you will get a different answer depending on whether you wiggle to one side of the pole or the other - no matter how small your make your wiggle - can you see why ? Gandalf61 (talk) 11:53, 17 April 2008 (UTC)

Understanding bases, and expanding them

Hello, all. I've read up on 1089 (number), which filled me with joy. I've been attempting to find the pattern between +1 digit answers, and it so far evades me. I'd be happy if anyone could lend a hand with that, perhaps some sumsign expression for all digit values 2 (99) to 10 (1,099,989,000)?

Anyway, this had me thinking about something else. Without thinking (contrary to my initial statement, I know, but the thinking follows from what ensued), I put the digit listings up for order. 1, 2, 3-... 10, 11, 12. Wait, this will not work beyond the 10th digit, because one is not allowed to repeat any digits! That made me think I might better be able to plot a pattern if I had more numbers. Hexadecimal would allow me this, but I'm not familiar with that, and it'd take me a long time. But wait! My computer should do these calculations in any photo editor, should it not? What's the base that computer graphics use? 81.93.102.185 (talk) 15:09, 17 April 2008 (UTC)

Computers use binary, or base 2, so I expect graphics processing is done in binary too. When you see computer stuff done in hex, that's just to make it easier for humans - since 16 is a power of 2, you can covert from binary to hex (and back) very easily. Each block of four bits (binary digits) corresponds to precisely one hexadecimal digit (so you can convert 4 bits at a time and don't need to look at the whole number at once as you would when converting to decimal). The computer uses binary, but converts to hex when displaying it for humans, since the numbers are shorter than way. --Tango (talk) 16:07, 17 April 2008 (UTC)

Also, stuff tends to be converted to hex (rather than base 32 or something) because 4 divides 8, and most things are organized into groups of 8 bits (=1 byte). This makes it a clear conversion of two hex digits into 8 binary digits. You can also take a look at wikipedia's article on bases. Someletters<Talk> 00:38, 18 April 2008 (UTC)

I would say the main reason for not using base 32 is because that's too many digits to remember and would confuse people, but the fact that it wouldn't work well to express a byte is also significant (but we could just use 10-bit bytes - I'm not sure having a power of two bits is essential) 16 is quite close to 10, so you only have a handful of extra digits. It would be great if you could express a byte as a single digit, but it would be more confusing that expressing it as two. You can't use octal because there wouldn't be an integer number of digits per byte, and using base 4 would be pretty pointless, it's not much better than binary. Base 16 is the smallest useful base. --Tango (talk) 15:50, 18 April 2008 (UTC)

"Power of two bits not essential . . " absolutely right, the UK police national computer installed in the 1970's used 20-bit words. However, 8-bit bytes and 16-bit words are pretty much the de facto standard everywhere now in the IT world so you would be swimming against the tide to do anything different now. "Base 16 is the smallest useful base" have to take issue with that, perhaps you meant "largest ueful". Base 10 is smaller and as that is used throughout the mathematics community I can only assume you find it useful. Base 12 was very popular in the commercial world (because it is abundant and therefore good for packing things in boxes) until the metrication nazis killed it off - that is smaller than 16 too. If you did mean largest, I would take issue with that too - at one time base 60 was the standard throughout the civilised world. That was also found rather useful, possibly for the same reason as 12. SpinningSpark 00:40, 20 April 2008 (UTC)

Sorry, I should have been clearer - I meant useful in computing, which restricts you to powers of two. Base 4 is so small you might as well just use binary, base 32 is too big to remember all the digits, which leaves bases 8 and 16 - octal and hexadecimal, and those are the two bases used in computing. What I said before isn't quite accurate, since base 8 is useful. I guess hex is chosen over octal in most uses because we use 8-bit bytes - if we used 6-bit bytes, we would do everything in octal. --Tango (talk) 00:52, 20 April 2008 (UTC)

On a related note, you mention "this will not work beyond the 10th digit, because one is not allowed to repeat any digits!" What's important to keep in mind here is what, precisely, is meant by "digit" in this context. It's simply a unique symbol agreed to be representative of something. For hexadecimal, the general agreement is 0123456789abcdef, with the understanding that "a" means base-10 "10". However, you could define a-f to instead be α-ζ if you preferred the Greek alphabet, and so long as the mapping is understood, those are perfectly valid digits. On the other hand, when you discuss binary, the symbol "2" is no longer a digit -- it's just a squiggle that can't be interpreted by the rules in question. So it's not a case of running out of digits but just understanding what the symbols represent. — Lomn 15:40, 18 April 2008 (UTC)

If you want to calculate in base 16 and you are using MS Windows then "Calculator" is a better bet than "Photoshop". You have to take it out of its standard view and put it in scientific view before you will see what you need. Once you have done that you can switch between hex and dec with one click. If you don't have "Calculator" then try this google search [1] to give you lots of on-line hex calculators. Also, if you have Excel, that can be persuaded to work in hex, but I seem to remember that it is an add-in that you have to load from the original disks. SpinningSpark 00:50, 20 April 2008 (UTC)

Calculus

Given the maclaurin series of f(x)

1/2! - x²/4! + x^4/6! - x^6/8! + ... + [(-1)^n x^2n]/(2n+2)!

How do I write this in terms of a familiar function, without series?

Thanks Thermophylae (talk) 17:37, 17 April 2008 (UTC)Thermopylae

Well, the standard series you are likely to know are for the exponential function, sine and cosine. Have a look at some of their series, and see which seems most like yours, and what the relation is. -mattbuck (Talk) 18:28, 17 April 2008 (UTC)

1+x²·f(x)=?. Bo Jacoby (talk) 18:05, 18 April 2008 (UTC).

Looks like the integral of some function... --wj32 ^t/c 03:55, 19 April 2008 (UTC)

OK, it isn't. But heres a clue: it's (1 - something) / (something else). I used Mathematica. --wj32 ^t/c 03:59, 19 April 2008 (UTC)

I recommend using Bo Jacoby's hint - calculate that series and see if you recognise anything. --Tango (talk) 14:27, 19 April 2008 (UTC)