What is the calculation used to place a cubic spline curve through 4 data points.?
459305
Suppose, we take 4 points:\nA (x_1, y_1)\nB (x_2, y_2)\nC (x_3, y_3)\nD (x_4, y_4)\nand a cubic spline y = ax^3 + bx^2 + cx + d\n\nTo find spline going through these points, this linear system must be solved (a,b,c,d are variables):\n\na*x_1^3 + b*x_1^2 + c*x_1 + d = 0\na*x_2^3 + b*x_2^2 + c*x_2 + d = 0\na*x_3^3 + b*x_3^2 + c*x_3 + d = 0\na*x_4^3 + b*x_4^2 + c*x_4 + d = 0
538371
http://en.wikipedia.org/wiki/Cubic_equation
976808
c=9\ns=28/15\nr=10\ny=-7/4\nfor the rest, i'd assume you're supposed to graph them need two points to graph lines so\n(-1,13)and(0,61,3)\n(0,-14)(14,0)\ndon't know what you mean by 7+1/4m...doesn't tell enough\n(0,5)(2.5,0)\n(0,5)(5/2,0)\n(1,22/6)(22/6,1)\n(0,7)(-21,0)
945415
On the off chance you want to know the steps, rather than a specific regression computer application, here's what you do:\n\nStep 1: collect data.  A minimum of two sets of data (like house prices and interest rates).  So, in this case, a single data "point" (p) would have p(price,rate) as the two sets of data making up each point.\n\nStep 2: after collecting a sufficient number of data points (more than 30 is good for better results), designate one set of data the "dependent" variable (e.g., y) and the other sets the "independent" variables (e.g., x1, x2, etc.)\n\nStep 3: set up a general linear equation (for linear regression) in the form of y = mx + b or, more generally, y = m1*x1 + m2*x2 + ... + mN xN + b, where each x term represents a set of independent data, y is the dependent data set, and b is a constant called the intercept.\n\nStep 4: enter the dependent and independent data sets into your regression tool (I use Excel's LINEST function for example).  Make sure you enter your dependent data into the y part of your regression tool and the x's, your independent data, into the independent data slots of the tool.  Otherwise, you'll get meaningless output from the tool\n\nStep 5: run the regression tool with your data entered. \n\nStep 6: note the output, which typically includes the mN coefficients for each x data set, the b intercept, R^2 the coefficient of determination, and measures of spread or uncertainty for each of the coeffcients and the intercept.
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...
751380
Since we're reflecting along a line parallel to the X-axis, the X coordinates of these points doesn't change.\n\nNew_Ay = Old_Ay + (4 - Old_Ay) * 2 = 3 + (4 - 3) * 2 = 5\nNew_By = Old_By + (4 - Old_By) * 2 = 5 + (4 - 5) * 2 = 3\nNew_Cy = Old_Cy + (4 - Old_Cy) * 2 = 6 + (4 - 6) * 2 = 2\n\nA = (2,5)\nB = (0,3)\nC = (5,2)
797690
convert to y=mx+b\nyou get\ny=4x-17\nslope =4
789559
Ten numbers per line, add 'em up and divide by 10.\n\nYou should get these means for the four sets:\n\n0.138, 0.172, 0.184, 0.156\n\nBe careful, since not all the numbers may show up on Yahoo unless you put your mouse pointer on the line.
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
498194
http://www.exploratorium.edu/baseball/curve.html
663533
In applied math, curve stitching (via spline interpolation, for example) is often used to approximate an original curve that is hard to calculate with with a manufactured curve that is easy to calculate. This makes it easier to see what the mathematical traits of the original curve might be without having to calculate the entire original curve.
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
478314
If you have a numerical approximation, your integration can not be more accurate than approximate, right?  And that may be close enough.  Integration is often the process of finding the area (integrating) under a curve, although the "curve" may be linear.  If you can draw your curve on a graph where for every value of x there is a single value of y, you can divide the graph into vertical slices where the bottom of each slice is the x axis and the top of each slice is a short horizontal line at the "eyeball" average height of the y axis in the slice.  If you divide the x axis into say five slices you then calculate the area in the five slices (A1 = x1 x y1, A2 = etc.).  You can increase the accuracy by using say 10 or 20 slices.  Of course calculus uses an infinite number of slices rather than approximation.\n\nAs a simple example, you can plot a "curve" of velocity vs time, where the velocity is Y and time is X.   In case one, velocity is constant (60 mph) and time increases to say 5 hours.  Y is a straight horizontal line at Y = 60 MPH from X = 0 to X = 5 Hr.  The "area" under the curve is a rectangle; 60 MPH x 5 Hr = 300 Miles.  So the integration represents distance traveled.\n\nIf in case two velocity increased linearly from Y = 0 to Y = 60 as time increased from X = 0 to X = 5, the area under the curve would be a triangle and can be calculated as 1/2 base time height or 1/2 (5 - 0) x (60 - 0) = 150 (miles).\n\nWhen the function is a real curve, you still need to estimate the area under the curve (with slices?) to integrate it.  You could have used slices in the above two examples to get an approximate answer but calculating area of a rectangle or triangle was easier and more accurate.  Good luck.
922643
It depends on the graph. If the values of each notch differ by x, then the interval is x.\n\nJust look at two numbers on adjacent notches, and subtract.
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
646014
I recommend "Unit 6: B-spline Curves" of this course:\nhttp://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/\n\nand in addition the learning software "DesignMentor V2: Curve system":\nhttp://www.cs.mtu.edu/~shene/NSF-2/DM2-BETA/index.html\n\nIt helps a lot to get an idea of base functions and partition of unity.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.