Wikipedia:Reference desk/Archives/Mathematics/2007 May 21

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May 21

What do I call this maths thing?

What do I call this maths thing, and where do I find it, what does it mean, and how does one get to the answer? Also, if there are any articles relating to this thing, can you please provide them? $5+{\ 14 \over {\ 5+{\ 14 \over {5\ +{\ 14 \over {5\ +...}}}}}}$ ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 19:05, 21 May 2007 (UTC)

I believe Continued fraction will have all the answers. -- Meni Rosenfeld (talk) 19:38, 21 May 2007 (UTC)

Thanks alot! I guess I should sharpen up my English skills lol :P ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 19:45, 21 May 2007 (UTC)

Although this thing is in fact a generalized continued fraction. The statement "... are precisely ..." about periodic continued fractions in the article Continued fraction also holds for the generalized ones if you make a small modification: omit the word "irrational" from that statement. In the generalized case you still get solutions of quadratic equations, but they may be rational. The articles do not actually say how to obtain the exact answer. It is rather simple, though. Call the whole thing x, so

$x=5+{\ 14 \over {\ 5+{\ 14 \over {5\ +...}}}}\,.$

Note that the (infinite) expression for x reappears as the denominator of the fraction whose numerator equals 14, so we can replace that denominator by x without change of meaning. This means that whatever the value of x is (assuming it has a value; but all continued fractions do), x satisfies

$x=5+{\ 14 \over x}\,.$

If you simplify this, you end up with a quadratic equation. It has two roots, but only one can be the right answer. It is not hard to figure out which one.  --Lambiam^Talk 21:54, 21 May 2007 (UTC)

Sorry, another question

Is it mathematically possible to determine k, if (x^3 - 7x^2 + kx + 13) / (x - 3) gives a remainder of 4? Isn't the value of x needed to find k? I've checked Polynomial remainder theorem and Factor theorem but I can't seem to find anything there. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 20:51, 21 May 2007 (UTC)

What do you get when you apply the polynomial remainder theorem to find the remainder of x³ − 7x² + kx + 13 divided by x − 3?  --Lambiam^Talk 21:23, 21 May 2007 (UTC)

This is polynomial long division, not regular division. — Daniel 21:26, 21 May 2007 (UTC)

What a fun question. Working with integer remainders, recall that removing all multiples of divisor d from n leaves a number less than d. Similarly, when working with polynomials the remainder must have degree less than that of the divisor. Here the divisor has degree one, so the remainder must be a constant, independent of x. Of course, in this case the "constant" will be a symbolic expression involving k. If the abstractions are too confusing, try assigning k numeric values from 1 to 5 and calculate the remainder each time; chances are you will quickly understand. --KSmrq^T 21:43, 21 May 2007 (UTC)

[Edit conflict] You may be confusing two different things: The remainder of division of the integers $x^{3}-7x^{2}+kx+13$ and $x-3$, for some unspecified but fixed integer value of x, as opposed to the remainder of division of the polynomials $x^{3}-7x^{2}+kx+13$ and $x-3$. The latter is discussed in Polynomial remainder theorem, and Lambiam hints at how to solve the problem. In case you mean the former, I suppose the problem is indeed unsolvable. -- Meni Rosenfeld (talk) 21:29, 21 May 2007 (UTC)