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September 17

Functions

I'm given a graph, however I am confused about how to solve something like $2f(x+3)=52$ 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)[reply]

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)[reply]

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)[reply]

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)[reply]

differential equation

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)[reply]

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

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

combinations and permutaions

what is value of nC3 —Preceding unsigned comment added by 59.184.240.35 (talk) 13:13, 17 September 2008 (UTC)[reply]

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

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 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)[reply]