solve an implicit function

Question

I have seen that an example of implicit function that can be solved only numerically is (solving for x knowing y): $$ \sin(x) = y\cdot x $$ I was wondering if the following function can also be solved only numerically: $$ x=\sqrt{x+1} + y $$ For me would be helpful to find a closed form solution, maybe using Mathematica, WxMaxima or any other software.

Thank you.

Just to add the maxima code: maxima workspace

While this is the equation I am trying to solve for $\beta$ (for which I thought that the problem was the root): $$ -((2*\beta^2+2)*d_x*m_x*\sqrt{(1-\theta^2)}+((\beta^2+1)*m_x^2+(-\beta^2-1)*m_x^2)*\theta^2+\sqrt{(\beta^2+1)}*((2*m_x*rho_1-2*m_x*p_z+2*\beta*d_y*m_x)*\theta-2*\beta*m_y*rho_1+2*\beta*m_y*p_z+2*d_y*m_y)+ (-\beta^2-1)*rho_1^2+(2*\beta^2+2)*p_z*rho_1+(-\beta^2-1)*p_z^2+(-\beta^2-1)*m_x^2+(-\beta^2-1)*m_y^2+(\beta^2+1)*L_r^2+(-\beta^2-1)*d_y^2+(-\beta^2-1)*d_x^2)/(\beta^2+1)=0 $$

Answer 1

Doing what the comments advice:

$$x^2-2xy+y^2=x+1\implies x^2-(2y+1)x+y^2-1=0$$

and this is a quadratic in $\;x\;$ :

$$\Delta=(2y+1)^2-4(y^2-1)=4y+5\implies x_{1,2}=\frac{(2y+1)\pm\sqrt\Delta}{2}$$

After that, substitute in the original equation and check what fits in.

Answer 2

To get rid of the square root, let $x+1=t^2$, with $t\ge0$. The equation now reads

$$t^2-t-y-1=0.$$

It has the solutions

$$t=\frac{1\pm\sqrt{1+4(y+1)}}2$$ which are defined for $4y+5\ge0$.

We have $t\ge0$ with the plus sign, or with the minus sign and $y\le-1$.

For these values,

$$x=t^2-1=\left(\frac{1\pm\sqrt{1+4(y+1)}}2\right)^2-1.$$

Answer 3

I deleted my last post because the calculation were rather cumbersome and I think it also contained some errors.

We start with this equation:

$$(-\left(2\,\beta^2+2\right)\,{\it d_x}\,{\it m_x}\,\sqrt{1-\vartheta ^2}-\left(\left(\beta^2+1\right)\,{\it m_x}^2+\left(-\beta^2-1\right) \,{\it m_x}^2\right)\,\vartheta^2-\sqrt{\beta^2+1}\,\left(\left(2\, {\it m_x}\,{\it \rho_1}-2\,{\it m_x}\,{\it p_z}+2\,\beta\,{\it d_y}\, {\it m_x}\right)\,\vartheta-2\,\beta\,{\it m_y}\,{\it \rho_1}+2\,\beta \,{\it m_y}\,{\it p_z}+2\,{\it d_y}\,{\it m_y}\right)-\left(-\beta^2-1 \right)\,{\it \rho_1}^2-\left(2\,\beta^2+2\right)\,{\it p_z}\, {\it \rho_1}-\left(-\beta^2-1\right)\,{\it p_z}^2-\left(-\beta^2-1 \right)\,{\it m_y}^2-\left(-\beta^2-1\right)\,{\it m_x}^2-\left(-\beta ^2-1\right)\,{\it d_y}^2-\left(-\beta^2-1\right)\,{\it d_x}^2-{\it L_r} ^2\,\left(\beta^2+1\right))/{(\beta^2+1)}=0$$

(%i1) eq:(-((2*beta^2+2)*dx*mx*sqrt(1-theta^2)
         +((beta^2+1)*mx^2+((-beta^2)-1)*mx^2)*theta^2
         +sqrt(beta^2+1)*((2*mx*rho1-2*mx*pz+2*beta*dy*mx)*theta
                         -2*beta*my*rho1+2*beta*my*pz+2*dy*my)
         +((-beta^2)-1)*rho1^2+(2*beta^2+2)*pz*rho1+((-beta^2)-1)*pz^2
         +((-beta^2)-1)*mx^2+((-beta^2)-1)*my^2+(beta^2+1)*Lr^2
         +((-beta^2)-1)*dy^2+((-beta^2)-1)*dx^2))
         /(beta^2+1)
          = 0
(%i2) eq1:eq
(%i3) eq2:solve(eq1,sqrt(beta^2+1))

Maxima transformed the equation to the form [sqrt(beta^2+1)=...]. Now we square the equation to get rid of this square root.

(%i4) eq3:eq2^2

Now we could solve this equation with repect to beta but we will get an erxpression that of about 30000 characters. That is useles for us. So we try to simplify the expression. Scanmap will map its function argument to all sub expression of the expression. First we try to factor any subespression then combine the coefficients of the same powers of beta to one coefficient (distribution law), e.g. u*beta + v*beta will be transformed to (u+v)*beta.

(%i5) eq4:scanmap(factor,eq3)
(%i6) eq5:scanmap(lambda([term],collectterms(term,beta)),eq4)

Te equation has now the following structure
Now all these coefficients of beta that contains parameters are replaced by a shortcut. We need three shortcuts and call them A, B and C.

(%i7) reveal(eq5,6)
                                         2       2
                              2       Sum (2) Sum (9)
(%o7)                    [beta  + 1 = ---------------]
                                              2
                                         4 Sum (4)
(%i8) part(reveal(eq5,6),1,2,1,2,1)
(%o8)                               Sum(9)
(%i9) t1:part(eq5,1,2,1,2,1)
(%i10) myvals:[C = t1]
(%i11) eq6:subst(C,ev(C,myvals),eq5)
(%i12) t2:part(eq6,1,2,2,2,1)
(%o12) beta (dy mx theta - my rho1 + my pz) + mx rho1 theta - mx pz theta
                                                                        + dy my
(%i13) t3:coeff(t2,beta,1)
(%o13)                   dy mx theta - my rho1 + my pz
(%i14) myvals:cons(A = t3,myvals)
(%i15) eq7:subst(A,ev(A,myvals),eq6)
(%i16) t4:part(eq7,1,2,2,2,1)
(%o16)           mx rho1 theta - mx pz theta + dy my + A beta
(%i17) t5:coeff(t4,beta,0)
(%o17)                mx rho1 theta - mx pz theta + dy my

these are the shortcuts, they are saved in the list myvals.

(%i18) myvals:cons(B = t5,myvals)
(%o18) [B = mx rho1 theta - mx pz theta + dy my,
A = dy mx theta - my rho1 + my pz, C =
                                              2                 2     2     2
2 dx mx sqrt(- (theta - 1) (theta + 1)) - rho1  + 2 pz rho1 - pz  - my  - mx
     2     2     2
 - dy  - dx  + Lr ]
(%i19) t6:ratsubst(B,ev(B,myvals),t4)

Now the equation has this structure and A,B and C are as displayed above.

(%i20) eq8:substpart(t6,eq6,1,2,2,2,1)


                                       2      2     2
                              2       C  (beta  + 1)
(%o20)                   [beta  + 1 = ---------------]
                                                    2
                                      4 (A beta + B)

If we substitute for A, B and C equation eq8 should be equal to eq3. We let Maxima check this. Obviously we didn't mek a mistake.

(%i21) is(equal(eq3,ev(eq8,myvals)))
(%o21)                               true

This are the solutions of this equation. The solutions %i and -%i can be ignored becuase these must be excluded becuase of the denominator of the original equation.

(%i22) t7:solve(eq8,beta)

We beautify the equation. This also removes the useless third and fourth solutions

(%i23) eq9:eq8/lhs(eq8[1])
(%i24) eq10:eq9*denom(rhs(eq9[1]))

Here is the equation and its two solutions. A, B and C are as defined above.

(%i25) eq11:eq10-rhs(eq10[1])
                                   2    2      2
(%o25)              [4 (A beta + B)  - C  (beta  + 1) = 0]
(%i26) t8:solve(eq11,beta)
                          2       2      2
               C sqrt((- C ) + 4 B  + 4 A ) + 4 A B
(%o26) [beta = ------------------------------------, 
                             2      2
                            C  - 4 A
                                                     2       2      2
                                          C sqrt((- C ) + 4 B  + 4 A ) - 4 A B
                                 beta = - ------------------------------------]
                                                        2      2
                                                       C  - 4 A

We now cannot check the equalities of this and the firs equation but we can check if the solutions of this equations are the same solutions as before. And this is true.

(%i27) is([t7[1],t7[2]] = t8)
(%o27)                               true

Shortcuts: $$A={\it d_y}\,{\it m_x}\,\vartheta-{\it m_y}\,{\it \rho_1}+{\it m_y}\, {\it p_z}$$ $$B={\it m_x}\,{\it \rho_1}\,\vartheta-{\it m_x}\,{\it p_z}\,\vartheta+ {\it d_y}\,{\it m_y}$$ $$C=2\,{\it d_x}\,{\it m_x}\,\sqrt{1-\vartheta^2}-{\it \rho_1}^2+2\, {\it p_z}\,{\it \rho_1}-{\it p_z}^2-{\it m_y}^2-{\it m_x}^2-{\it d_y}^2- {\it d_x}^2+{\it L_r}^2$$

Equation: $$4\,\left(A\,\beta+B\right)^2-C^2\,\left(\beta^2+1\right)=0$$

Solutions:

$$\beta_1={{C\,\sqrt{-C^2+4\,B^2+4\,A^2}+4\,A\,B}\over{C^2-4\,A^2}}$$ $$\beta_2=-{{C\,\sqrt{-C^2+4\,B^2+4\,A^2}-4\,A\,B}\over{C^2-4\,A^2}}$$