I need to know how to write a function to solve a simple linear equation like 2x +1 = 5. How would one do this? If anyone can show some code or point me to a site, it would be much appreciated.
How is the equation entered? In text-form? As coefficients?
ax + b = c of course has the solution x = (c - b) / a.
For parsing, you could use regular expressions.
Parse the expression into a simple tree, then use basic math to solve it. I'd start by converting it to postfix notation and then evaluating that.
Related
I am trying to integrate a Heaviside theta function with two signs inside and Mathematica won't give me a solution. Is there any way of improving the approach before just acknowledging that Mathematica cannot integrate it?
Some things you do really worry me.
You use X as a variable and as the name of a function. I've changed those to X and Xf.
You use ω as a variable and as the name of a function. I've changed those to ω and ωf.
You use = and not := in your function definitions. I've changed that.
With those changes I have
Xf[s1_,s2_,α_]:=(1+s2(-2+α)-α+s1(-1+2s2+α))/((-1+2 s1)(-1+2s2));
ωf[s1_,s2_,α_]:=((1-s2+s1(-3+4s2))(1+s2(-2+α)-α+s1(-1+2s2+α)))/((-1+2s1)(-1+s1+s2)(-1+2s2));
λ1[s1_,s2_,α_,X_,ω_]:=1/2(s1-2s2-3s1 X+3s2 X-α+s1 α+s2 α+ω-s1 ω-s2 ω-
1/2Sqrt[(4(α-ω+s2(2-3X-α-ω)+s1(-1+3X-α+ω))^2+8(2-5X+4X^2-2α+2X α+
s1(-2+4s2(1-2X)^2+7X-8X^2+2α-2X α-ω)+ω-s2(4-11X+8X^2-2α+2X α+ω)))]);
λ2[s1_,s2_,α_,X_,ω_]:=1/2(s1-2s2-3s1 X+3s2 X-α+s1 α+s2 α+ω-s1 ω-s2 ω+
1/2Sqrt[(4(α-ω+s2(2-3X-α-ω)+s1(-1+3X-α+ω))^2+8(2-5X+4X^2-2α+2X α+
s1(-2+4s2(1-2X)^2+7X-8X^2+2α-2X α-ω)+ω-s2(4-11X+8X^2-2α+2X α+ω)))]);
λ1simp[s1_,s2_,α_]:=λ1[s1,s2,α,Xf[s1,s2,α],ωf[s1,s2,α]];
λ2simp[s1_,s2_,α_]:=λ2[s1,s2,α,Xf[s1,s2,α],ωf[s1,s2,α]];
fint[s1_,s2_]:=HeavisideTheta[Sign[-λ1simp[s1,s2,α]]*Sign[-λ2simp[s1,s2,α]]];
Please check that very carefully to see if I've made any mistakes.
Now I want to look at your integrand and see what Mathematica sees.
Simplify[fint[s1,s2],1/2<=s1<=1&&1/2<=s2<=s1]
And it responds with
HeavisideTheta[CompexInfinity] 2s1==1||2s2==1||s1+s2==1
HeavisideTheta[True Sign[...]*Sign[...]]
so it looks like your integrand is blowing up at the boundary.
I check that with
Simplify[fint[1/2,s2]]
or
Simplify[fint[s1,1/2]]
and it responds with 1/0 and Indeterminate and HeavisideTheta[Indeterminate]
When it isn't at the boundary, for example
Simplify[fint[3/4,3/4]]
it returns
HeavisideTheta[Sign[5-4*Sqrt[7-28*α+25*α^2]]*Sign[5+4*Sqrt[7-28*α+25*α^2]]]
and that probably says that α is a free variable and we aren't able to determine the value of the Sign without more information.
So I think this is a strong hint where I would begin looking for why your integration is not simply completing.
If you are curious what that integrand looks like then try
α=1/4;
Plot3D[fint[s1,s2],{s1,1/2,1},{s2,1/2,s1}]
All I know is that a single integer will surely accept it.
The equation is like :
Ax^5 + Bx^3 + Cx^2 = D
I tried to brute force value of x , but was getting TLE , can I use an optimised binary search as I know only one root will be real?
You may want to search for Newton-Raphson's method which is known to quickly converge to solution with just a few iterations.
You're just asking to find the zeros of a function when you've been guaranteed that there's no more than one zero. To put it concretely, let's assume you have the following equation:
-15x^5 + 12x^3 - 203x^2 = -2.193113e+12
You could use the root-finding function from your favorite statistical software package to find the root. For instance, here's how you would do it with uniroot in R:
uniroot(function(x) -15*x^5 + 12*x^3 - 203*x^2 + 2.193113e+12, c(-1000, 1000))$root
# [1] 171
You could try typing this into Wolfram Alpha.
Solve[3x^5+4 x^3+5x^2==148,x]
I am trying to find the runtime of the equation;
T(n) = T(n-2) + n³.
When I am solving it I arrive at the summation T(n) = T(n-k) + Σk = 0,...,n/2(n-2k)³.
Solving that sum I get 1/8(n²)(n + 2)². Solving this I would get the runtime to be Θ(n⁴).
However, I think I did something wrong, does anyone have any ideas?
Why do you think that it is wrong? This equation is clearly Theta(n^4)
The more detailed solution can be obtained from WolframALpha (did you know it solves recurrence equations?)
https://www.wolframalpha.com/input/?i=T%28n%29%3DT%28n-2%29%2Bn%5E3
You can also add some border cases, like T(0)=T(1)=1
https://www.wolframalpha.com/input/?i=T%28n%29%3DT%28n-2%29%2Bn%5E3%2C+T%281%29%3D1%2C+T%282%29%3D1
and finally: asymptotic plot, showing that it truly behaves like n^4 function
Here is an attempt to show your recursive recursive recurrence with steps:
With WolframAlpha engine solving the summation.
how to write a formula like
v_r (t)=∑_(n=0)^(N-1)▒〖A_r (L_2-L_1 ) e^j(ω_c t-4π/λ (R+υt+L_(1+L_2 )/2 cos〖(θ)sin(ω_r t+2πn/N)))〗 ┤) sinc(4π/λ-L_(2-L_1 )/2 cos(θ) sin(ω_r t+2πn/N))〗
in c#?
You have to convert the formula to something the compiler recognizes.
To it's equivalent using the a combination of basic algebra and the Math class like so:
p = rho*R*T + (B_0*R*T-A_0-((C_0) / (T*T))+((E_0) / (Math.Pow(T, 4))))*rho*rho +
(b*R*T-a-((d) / (T)))*Math.Pow(rho, 3) +
alpha*(a+((d) / (t)))*Math.Pow(rho, 6) +
((c*Math.Pow(rho, 3)) / (T*T))*(1+gamma*rho*rho)*Math.Exp(-gamma*rho*rho);
Example taken from: Converting Math Equations in C#
Well, first you have to figure out what all those symbols mean. I see the sigma which usually indicates sum-of, with ∑_(n=0)^(N-1) probably translating to:
N-1
∑
n=0
This generally means the sum of the following expression where n varies from 0 to N-1. So I gather you'd need a loop there.
The expression to be calculated within that loop consists of a lof of trigonometric-type functions involving π, θ, sin and cos, and the little known sinc which I assume is a typo :-)
The bottom line is that you need to understand the current expression before you can think about converting it to another form like a C# program. Short of knowing where it came from, or a little bit of context, we probably can't help you that much though there's always a possibility that we have a savant/genius here that will recognise that formula off the top of their head.
I'm trying to solve a set of symbolic equations in maxima.
I'm using "algsys" comand, and it works, but the answer is depended on "%r1".
I dont understand what %r1 is, and how can I get a explicit expression?
maxima code:
(%i4) eq1:diff(diff(Lag,diff(u(t),t)),t)-diff(Lag,u(t)),ratsimp;
(%i5) eq2:diff(diff(Lag,diff(v(t),t)),t)-diff(Lag,v(t)),ratsimp;
(%i6) eq3:diff(diff(Lag,diff(theta(t),t)),t)-diff(Lag,theta(t))-Tm,ratsimp;
(%i7) A:algsys([eq1, eq2, eq3], [diff(u(t),t,2),diff(v(t),t,2),diff(theta(t),t,2)]);
PS: cant post picture cause it's my first question...
FWIW %r1 is simply names for roots introduced by solve and algsys.
See http://maxima.sourceforge.net/docs/manual/en/maxima_20.html