Wikipedia:Reference desk/Archives/Mathematics/2008 September 17#differential equation

I'm given a graph, however I am confused about how to solve something like $2\; f(x+3)=52$ algebraically for x. Could someone please point me in the right direction. (By the way, this is a homework question.) I have to show my work, or I would just use the graph. TIA, 67.54.224.199 (talk) 01:45, 17 September 2008 (UTC)

:If you are given an actual formula for f, then you can solve algebraically, starting with appropriate substitution. For example, if you were given f(x) = x^2, then f(x+3) = (x+3)^2 and from that you then solve 2 (x+3)^2 = 52. If you are only given a graph, then you would need to first use the graph to find an approximate solution to 2 f(x') = 52, and then using that approximate solution to solve x' = x + 3 for x. Confusing Manifestation(Say hi!) 03:50, 17 September 2008 (UTC)

::In more details, locate the intersections of your graph and the horizontal line y=26 (52/2) to find x' = x+3. Then x = x'-3 are your solutions if there are more than one point in the intersection. twma 10:08, 17 September 2008 (UTC)

Need x so that: f(x+3) = 26. Therefore, you find x' such that f(x') = 26 and subtract 3 from x' to get the desired x (since f(x+3) = f(x'-3+3) = f(x') = 26 as desired).

The most important thing to remember is that f(x-a) translates the graph a units to the right (to get the same value of f(x), you need to add 'a' to the x-value). Similarly, f(x+a) translates the graph 'a' units to the left (to get the same value of f(x), you need to subtract a from the x-value so the graph must be translated 'a' units left). Similarly, -f(x) reflects f about x-axis and f(-x) reflects f about y-axis. In general a*f(x) dilates graph parallel to y-axis; scale factor 1/a. Similarly f(ax) dilates graph parallel to x-axis scale factor 1/a (since you need 1 a^(th) the same value of x to get the original f(x) so the graph must shrink horizontally) —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:03, 17 September 2008 (UTC)

Hi, is there a solution to xy" + y' - y = 0 that uses a finite set of well known functions?(Not the power series solution--I got the d.e. as something a certain power series satisfied and I want to express the function repped by the p.s. in terms of well known functions if possible. The function is y = the sum of (x^n)/[(n!)^2] from n=0 to infinity, which I got from the linear operator that sends the p.s. for 1/(1-x) to the one for e^x. When you do the same on the p.s. for e^x, you get y.)Thanks in advance.Rich (talk) 02:54, 17 September 2008 (UTC)

:See Bessel-Clifford function with n=0. Dmcq (talk) 09:49, 17 September 2008 (UTC)

::good referral, thank you.130.86.14.25 (talk) 05:10, 19 September 2008 (UTC)

what is value of nC3

class="autosigned">—Preceding unsigned comment added by 59.184.240.35 (talk) 13:13, 17 September 2008 (UTC)

:See Combination. --Tango (talk) 13:19, 17 September 2008 (UTC)

:Use the definition of binomial coefficient, substitute 3, and simplify. (Alternately, use the theorem that nCk is a polynomial of n of degree k for any fixed nonnegative integer k, compute it for a few small values, then interpolate. Alternately, compute for the first few values of n and look up in [http://www.research.att.com/~njas/sequences/ OEIS] for a formula. Alternately rewrite the binomial to a falling power and use the equation about falling powers and Stirling numbers to get the result in expanded forms rightaway.)

:If you don't know what binomial coefficients are in general then a good introduction might be the first few chapters of Graham–Knuth–Patashnik, Concrete mathematics. – b_jonas 15:15, 19 September 2008 (UTC)