I'm trying to solve x for any given cubic when y is known(not roots). Ex: 3x^3 + 4x^2 + 2x + 2 = y = 46 If I know y, how can I find x? I'm trying to create an equivalent pascal function of the Mathcad interp function. Any links or explanations are highly appreciated. MP ==== MP, I'm trying to solve x for any given cubic when y is known(not roots). Ex: 3x^3 + 4x^2 + 2x + 2 = y = 46 If I know y, how can I find x? This is still essentially a root problem. In your example, 3x^3+4x^2+2x+2=46 3x^3+4x^2+2x-44=0 Producing all three solutions to the cubic is a long-solved problem, but the formulas can get a little ugly. Check out http://mathworld.wolfram.com/CubicEquation.html . However, if you know a little bit about the roots of the cubic, you can save yourself a good deal of trouble. If you know their approximate values, then you can use Newton's method. Given the cubic polynomial p(x)=3x^3+4x^2+2x-44 (in your case), pick an value x_0 near a root. Then, let x_1 = x_0 - p(x_0) / p'(x_0) ... = ... x_(n+1) = x_n - p(x_0) / p'(x_0) Where p'(x) is the derivate of p(x). If you know that (at least) one of the roots is a rational, then you can test all the possible roots, find a rational root a, divide p(x) by (x-a), and then use the quadratic formula on the remaining polynomial. There are several more tricks, whose applicability depend on the specifics of the problem. Travis ==== 1st subtract 46 from both sides and get 3x^3 + 4x^2 + 2x - 44 = 0. I'd then check to see if there are any rational roots by using the rational root theorem. I'm trying to solve x for any given cubic when y is known(not roots). Ex: 3x^3 + 4x^2 + 2x + 2 = y = 46 If I know y, how can I find x? I'm trying to create an equivalent pascal function of the Mathcad > interp function. Any links or explanations are highly appreciated. > MP X-Admin: news@cs.com ==== >1st subtract 46 from both sides and get 3x^3 + 4x^2 + 2x - 44 = 0. I'd then >check to see if there are any rational roots by using the rational root >theorem. Synthetic division to look for a way to break the polynomial into simpler binomial and monomial factors? G C ==== I'm trying to solve x for any given cubic when y is known(not roots). Ex: 3x^3 + 4x^2 + 2x + 2 = y = 46 If I know y, how can I find x? I'm trying to create an equivalent pascal function of the Mathcad >interp function. Any links or explanations are highly appreciated. >MP > As others have said you rewrite your equation in the form x^3+ax^2+b^x+c=0 then you can use cardano's formula to find the roots, see http://lib-www.lanl.gov/numerical/bookcpdf/c5-6.pdf for a numerically stable way to do that. ==== I keep running into the same type of integral & I can't remember how to do them...any help is appreciated: Ex: sqrt(16-r^2) integral of (square root of quantity 16 minus r squared) ==== >I keep running into the same type of integral & I can't remember how to do >them...any help is appreciated: Ex: sqrt(16-r^2) integral of (square root of quantity 16 minus r squared) An arc sine is involved. Some calculus books cover this under trigonometric substitution, others under rational functions of sin and cos. For your specific one, my TI-89 says INT [ sqrt(16-r^2) dr ] = 8*arcsin(r/4) + (r/2)*sqrt(16-r^2) + C -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== JD- > I keep running into the same type of integral & I can't remember how to do > them...any help is appreciated: Ex: sqrt(16-r^2) integral of (square root of quantity 16 minus r squared) This is a classic trigonometric substitution problem. Let r = 4cos(t). Then, dr = -4sin(t)*dt Then, Integral[sqrt(16-r^2)*dr] =Integral[sqrt(16-(4cos(t))^2)*(-4sin(t)dt)] =-4*Integral[sqrt(16-16(cos(t))^2)*sin(t)dt] =-16*Integral[sqrt(1-(cos(t))^2)*sin(t)dt] =-16*Integral[(sin(t))^2*dt] The rest should be clear: evaluate the above integral (hint: (sin(t))^2=(1-cos(2t))/2), and then put your expression back into terms of r. (t=arccos(r/4)) Now, sqrt(a^2-r^2) is just the formula for the upper half of a circle of radius a centered on the origin. So, with a little common sense, you can ignore integration completely if the limits are nice. (For example, if you're integrating over [0,a], then your region is a quarter of a disk of radius a: therefore, the integral is (pi*a^2)/4. Travis ==== >I keep running into the same type of integral & I can't remember how to do >them...any help is appreciated: Ex: sqrt(16-r^2) integral of (square root of quantity 16 minus r squared) > See http://www.karlscalculus.org/calc11_4.html for trig substitution and also the rest of the site. ==== Can anyone please calculate the value of C showing working: E^19000000c = 2 I know it involves the logarithmic function on calculators but I can't get my calculator to do it.