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September 10

Query regarding divisibilty

We have got some basic ways to find out the divisibility of an integer(irrespective of its size) by 2,3.. til 10 (within 10). But is there any basic ways to get it applied for divisibility by 7. —Preceding unsigned comment added by Bbsxyz (talk • contribs) 14:44, 10 September 2008 (UTC)[reply]

There's nothing very easy. The equivalent of the usual test for divisibility for 3 is that if you take the units digit of your number, add 3 times the tens digit, add 2 times the hundreds digit, subtract the thousands digit, subtract three times the tenthousands digit, subtract 2 times the hundredthousands digit, add the millions digit, and so on (repeating the pattern 1, 3, 2, -1, -3, -2) then the result of this sum is divisible by 7 if and only if the original number was. That probably isn't any quicker by hand than just long division by 7 though. Algebraist 14:50, 10 September 2008 (UTC)[reply]

See Divisibility rule#Divisibility by 7. Gandalf61 (talk) 14:54, 10 September 2008 (UTC)[reply]

That's correct, and fast. A slower method, but easier to remember is to expand the 'number value' taking 3 instead of 10 as the system base.

Say a number 2175 = 2×10³ + 1×10² + 7×10¹ + 5×10⁰

Calculate its 3–base value: 2175 → 2×3³ + 1×3² + 7×3¹ + 5×3⁰ = 2×27 + 1×9 + 7×3 + 5×1 = 89

Iterate the method: 89 → 8×3¹ + 9×3⁰ = 8×3 + 9×1 = 33

33 → 3×3¹ + 3×3⁰ = 3×3 + 3×1 = 12

12 → 1×3¹ + 2×3⁰ = 1×3 + 2×1 = 5

Number 5 is not divisible by 7, so the initial number 2175 is not, either. Also 89, 33 and 12 are not divisible — all of them give the remainder of 5.

The coefficients given by Algebraist and Gandalf above are simply equivalents of powers of 3 with respect to divisibility by 7 (we say 'congruent modulo 7', see Modular arithmetic), chosen as close to zero as possible, so that numbers calculated faster converge to (-6, 6) interval. --CiaPan (talk) 07:19, 11 September 2008 (UTC)[reply]

What are the rules of treating differentials?

I learned calculus from a Stewart book which looks down on using dummy variables as actual terms like dv = a dt

I am wondering if someone can point me to a page (here or on the internet) that outlines the rules for how engineers and physicists treat the differentials as meaningful terms. My Stewart book said that the dt that appears after the integrand is only to remind us how Leibniz developed them from the concept of Reimann sums.

I'm specifically looking at

$dv=-v_{rel}{\frac {dM}{M}}$

$\int _{v_{i}}^{v_{f}}dv=-v_{rel}\int _{M_{i}}^{M_{f}}{\frac {dM}{M}}$

I don't understand how in the top equation, I can integrate both sides between different limits of integration? In my newbie understanding, that is like dividing one side of an equation by 12 candelas, but dividing the other side by 13 pascals.

My second question, is changing the limits of integration from v_i and v_f to M_i and M_f. I understand Integration by substitution clearly as it is taught in my Stewart book. I could understand if both sides were integrated between limits of integration if the terms had the same dimensions (velocity and velocity) but then one of the integrals changed the integrand from a velocity to a mass (and in complying with the FTC-2 of the subsitution rule, also changing the limits v_i to M_i and v_f to M_f).

I guess my apprehension is worrying about if the left side and right side will balance.

Third question is 5 second easy one. When you evaluate a definite integral, does the answer always have the same dimension as the integrand multiplied by the dummy variable? This is a pattern I've noticed.

Thanks, I'm a contributor to the Differential article and other pages where I can help out with clearity and links and navigability. Also, I've read 7 articles trying to find my answer, and have tried looking online. Thanks in advance. My only 3 questions at the ref desk have all been about differentials. Sentriclecub (talk) 14:56, 10 September 2008 (UTC)[reply]

Here's my work so far...

$\int _{a}^{b}dv=(b-a)\cdot y$

and I also have this worked out

$-v_{rel}\int _{a}^{b}{\frac {dM}{M}}=-v_{rel}\cdot (\ln b-\ln a)=-v_{rel}\cdot \ln {\tfrac {b}{a}}$

So as you can see, I'm stuck on only seeing the dM and the dv as a dummy variable only, and want to use y = f(x) lingo that is in my engrams. —Preceding unsigned comment added by Sentriclecub (talk • contribs) 15:23, 10 September 2008 (UTC)[reply]

You can think (extremely unrigorously) of the "dv" or "dM" or whatever as being an infinitesimal change in v or M, or whatever, so dM is probably the mass of an infinitesimal amount of the item in question. You can manipulate them in much the same way as any other variable and it usually works just fine, but just don't let a mathematician see you doing it! (And sanity check your answers just in case you've found one of the few cases where it doesn't work.) As for adding limits on both sides, I think they need to be the same limit, just in terms of the appropriate variable (so your limits are probably the values at t=0 and the values at t=T, or something, you just use the velocity at those times on one side and the mass at those times on the other). And yes, the dimension of d(Something) is the same as the dimension of the something and integration just means adding lots of them together, so that doesn't change the dimension. --Tango (talk) 15:31, 10 September 2008 (UTC)[reply]

If you like to be rigorous, just treat dx and dy as new variables. The rules of algebra are the addition rule: d(x+y) = dx+dy, the multiplication rule: d(x·y) = x·dy+dx·y, the power rule: d(x^y) = y·x^y−1·dx+x^y·log(x)·dy, and the constant rule: dk = 0 if k is constant. Refer to differential algebra in case mathematicians object, as they sometimes do because they got scared by the unrigorous interpretation of differentials as infinitesimals. Note that the limits refer to the dummy variable behind the d. Explicitely:

\int _{a}^{b}dv=\int _{v=a}^{b}dv=\int _{x=a}^{b}dx=b-a

\int _{v=v_{i}}^{v_{f}}dv=-v_{rel}\int _{M=M_{i}}^{M_{f}}{\frac {dM}{M}}

Note that M_i=M(v_i). Bo Jacoby (talk) 16:52, 10 September 2008 (UTC).[reply]

Thanks so much for the help. I knew I understood it when you adduced Note that M_i=M(v_i) because I have spent the last 90 minutes, and my brainstorming was that a and b are points that have a lot of information. I conjectured a function that assigns a to v_i and a function that assigns a to M_i and then integrated this function from initial to final.

I reasoned that I am still only integrating from a to b and the only difference is that I'm now integrating between f(a) and f(b) a new function which assigns to each element f(c₁) in a set I to an element c₁ in a set II

I concluded that every value between a and b has a corresponding M_i and a corresponding v_i. Thus some value between a and b has a corresponding M_j and v can be thought of as a function of M since every M_j has only one corresponding v_j. Thanks very much! I bookmarked this great explanation from my userpage, and I tried a problem from my book and it was finally easy, clear, and simple. Sentriclecub (talk) 17:35, 10 September 2008 (UTC)[reply]

You've certainly got your work cut out to distinguish clearly between the various forms of differentials and derivatives. Offhand I can think of things like partial derivative, exterior differential, covariant derivative, never mind the boundary operator looking like a partial derivative. Using differentials is good I think but there is some problems especially with twice differentiating. In differential forms d² is zero, and for ordinary differentials y"dx² isn't the same as d(dy) which if one works it out would come to y"dx²+y'd(dx). Dmcq (talk) 18:04, 10 September 2008 (UTC)[reply]

Thanks dmcq for the input. I trust the reasoning behind Stewart's exclusion of this information from his formal calculus textbook. I purposely tried to avoid learning the stuff about shorthand differentials treating them as autonomous terms, but I only did so to make better sense of my physics textbook. Plus the public lectures from MIT's opencourseware, is taught by professor Walter Lewin who uses this treatment of differentials in upcoming electromagnetism lectures. Both these were a factor, but I wish I had the time-machine to go back to age 18 and majored in physics and not finance. I'm now halfway to 30 and have a lot to catch up! Sentriclecub (talk) 18:27, 10 September 2008 (UTC)[reply]

Ulp I thought I say the quote marks in what I wrote but obviously I didn't and they have turned into the thing to make italic writing. I'll fix that last line and write it here again. For ordinary differentials y"dx² isn't the same as d(dy) which if one works it out would come to y"dx²+y'd(dx). One has to resist the urge to use d²x here. There's also Lie derivatives if you're interested. Dmcq (talk) 18:52, 10 September 2008 (UTC)[reply]

Dmcq points at the difficulty in differentiation twice using differential notation. The clue here is that when x is an independent variable, then dx is constant, and so ddx=d²x is zero. So y=x² gives dy=d(x²)=2·x·dx and d²y=d(dy)=d(2·x·dx)=2·d(x·dx) =2·(dx·dx+x·d(dx))=2·(dx²+x·d²x) =2·dx² and d²y/dx²=2. Bo Jacoby (talk) 09:41, 11 September 2008 (UTC).[reply]

True if you can keep the business of what is independent clearly in mind. In fat you might like the full expression because simply by setting d² to zero so y is the independent variable you get a nice way of deriving d²x/dy² from the value of d²y/dx² Dmcq (talk) 18:38, 11 September 2008 (UTC)[reply]

Stewart is a fanatical adherent of a religion that holds that all calculus textbooks must be identical in all essentials (and you make yours brilliantly different by doing something really original such as putting the Mean Value Theorem in Section 4.2 instead of Section 4.1, which has been done by only 950 previous authors, or slightly changing the colors of the figures, and that although one must not attempt logical rigor, nonetheless the order in which one presents the topics must be what would make sense if one did. That religion forbids being honest about the Leibniz notation, and says one must tell students that they're no good and just historical curiosities. Michael Hardy (talk) 05:53, 14 September 2008 (UTC)[reply]

mathematician name for famous subset of whole #'s 1,2,3,5,8,13,21 ?

Trying to help my Daughter with no Text book —Preceding unsigned comment added by 24.215.67.178 (talk) 21:26, 10 September 2008 (UTC)[reply]

Fibonacci. -- The Anome (talk) 21:30, 10 September 2008 (UTC)[reply]

There is a web-site, the On-Line Encyclopedia of Integer Sequences (link), where you can type a sequence such as yours, and get it identified if it is in any way notable. typing in your numbers, returns

Fibonacci numbers: F(n) = F(n-1) + F(n-2), F(0) = 0, F(1) = 1, F(2) = 1,

Note, also, that the first two numbers of the sequence are missing in your question, it should be 0,1,1,2,3,5,8,13,21,... Except for these two first numbers, every number is the sum of the two preceding ones. See Fibonacci number. --NorwegianBlue^talk 23:44, 10 September 2008 (UTC)[reply]

Actually, even those two numbers are the sum of the two preceding ones. The Fibonacci sequence is easily extended in the negative direction. Fredrik Johansson 12:40, 12 September 2008 (UTC)[reply]