First doing by hand , to see where it goes wrong in the code : 
Step 3 is a vital one , but not easy to get , by knowing this 
A nightmare this 

# Step 1: Define original expression
expr := sqrt(4 - 2*y*(x + 1));

# Step 2: Expand the expression inside the root
expandedjd := expand(4 - 2*y*(x + 1));  # Gives -2 x y - 2 y + 4

# Step 3: Factor using symbolic simplification
factored := simplify(expandedjd, symbolic);  # Returns 2*(2 - y*(x + 1))

# Step 4: Apply the square root to the factored expression
root_factored := sqrt(factored);  # sqrt(2*(2 - y*(x + 1)))

# Step 5: Split the root into a product of square roots
split_root := sqrt(2)*sqrt(2 - y*(x + 1));

# Step 6: Verify equivalence
check := simplify(root_factored - split_root);  # Should return 0

Yes — for elliptic integrals of the second (and first) kind with real arguments, this is always true.
sin(x) never causes branching there; all branch information runs via cos(x). ( ai  ?) 
Turns out it has a very unique approach?
 

I don't know how to use all elliptic integrals to obtain the desired output by specifying conditions for the integral definitions.
That would need to be investigated, but is it now known for one elliptic integral, but not direct?

restart:
f := sqrt(1 + sin(x)^2);
F := int(f, x);
df := diff(F, x):
simplify(df - f);
simplify(df - f, symbolic);

restart;
f := sqrt(1 + sin(x)^2);
F := int(f, x);
F := simplify(F);
#F := simplify(F,symbolic);
df := diff(F, x);
assume(cos(x) > 0):
simplify(df - f);

piecewise(
  cos(x) > 0,  EllipticE(sin(x), I),
  cos(x) < 0, -EllipticE(sin(x), I)
);

restart:
f := sqrt(1 + sin(x)^2):
F := int(f, x);
df := diff(F, x);
simplify(df - f);  # ==> must be give  0 

Maybe too much ?
Showing a histogram plot for the 3 methods seperately is better ?
It is in build in maple ?

(1) >      >  restart; with(Student:-Calculus1): with(plots): f := x -> x^2; a := -2; b := 2; exact := int(f(x), x = a..b); # Compacte animatie animate(     n -> display(         # Drie methodes         RiemannSum(f(x), x = a..b, method = left, partition = trunc(n),                   output = plot, caption = "", boxoptions = [filled = [color = "LightBlue"]]),         RiemannSum(f(x), x = a..b, method = right, partition = trunc(n),                   output = plot, caption = "", boxoptions = [filled = [color = "LightCoral"]]),         RiemannSum(f(x), x = a..b, method = midpoint, partition = trunc(n),                   output = plot, caption = "", boxoptions = [filled = [color = "LightGreen"]]),                  # Functie         plot(f(x), x = a..b, color = "Black", thickness = 2),                  # Info         title = typeset("Riemann Methods: n=", trunc(n),                        ", Exact=", evalf[5](exact)),         view = [a..b, -0.2..4.5]     ),     [n],     n = 1..150,     frames = 25,     size = [1000, 400] ); >      >  restart; with(Student:-Calculus1): with(plots): f := x -> x^2; a := -2; b := 2; exact := evalf(int(f(x), x = a..b)); # CLEAR VISUALIZATION: Three separate plots side by side animate(     n -> display(         Array([             # COLUMN 1: LEFT ENDPOINT METHOD             display(                 RiemannSum(f(x), x = a..b, method = left, partition = trunc(n),                           output = plot, caption = "",                           boxoptions = [filled = [color = "Blue", transparency = 0.6]]),                 plot(f(x), x = a..b, color = "Black", thickness = 2),                 title = typeset("LEFT ENDPOINT METHOD\n",                                "Uses: f(x₍ᵢ₋₁₎) at LEFT of each interval\n",                                "Approximation: ",                                evalf[5](RiemannSum(f(x), x=a..b, method=left, partition=trunc(n)))),                 view = [a..b, -0.5..4.5]             ),                          # COLUMN 2: RIGHT ENDPOINT METHOD             display(                 RiemannSum(f(x), x = a..b, method = right, partition = trunc(n),                           output = plot, caption = "",                           boxoptions = [filled = [color = "Red", transparency = 0.6]]),                 plot(f(x), x = a..b, color = "Black", thickness = 2),                 title = typeset("RIGHT ENDPOINT METHOD\n",                                "Uses: f(xᵢ) at RIGHT of each interval\n",                                "Approximation: ",                                evalf[5](RiemannSum(f(x), x=a..b, method=right, partition=trunc(n)))),                 view = [a..b, -0.5..4.5]             ),                          # COLUMN 3: MIDPOINT METHOD             display(                 RiemannSum(f(x), x = a..b, method = midpoint, partition = trunc(n),                           output = plot, caption = "",                           boxoptions = [filled = [color = "Green", transparency = 0.6]]),                 plot(f(x), x = a..b, color = "Black", thickness = 2),                 title = typeset("MIDPOINT METHOD\n",                                "Uses: f((x₍ᵢ₋₁₎+xᵢ)/2) at MIDDLE of each interval\n",                                "Approximation: ",                                evalf[5](RiemannSum(f(x), x=a..b, method=midpoint, partition=trunc(n)))),                 view = [a..b, -0.5..4.5]             )         ]),         title = typeset("VISUAL COMPARISON OF RIEMANN SUM METHODS\n",                        "f(x) = x² on [", a, ", ", b, "] | n = ", trunc(n),                        " partitions | Δx = ", evalf[5]((b-a)/trunc(n)),                        " | Exact integral = ", evalf[5](exact)),         size = [1200, 450]     ),     [n],     n = 1..50,     frames = 25,     paraminfo = false ); >    >

Download riemann_som_mprimes_DEF_22-12-2025.mw

why_my_solution_not_stisfy_th_e_pdes_mprimes_21-12-2025.mw

b13_vraag_mprimes_15-12-20225.mw

@Alfred_F 

THE POSSIBLE 'AHA MOMENT':
============================

Lipschitz was probably struggling with:
How do you prove convergence without assumptions about f'?

INSIGHT: "What if I REVERSE the MEAN VALUE THEOREM?
Instead of: IF f is differentiable, THEN |f(u)-f(v)| ≤ sup|f'|·|u-v|
Take as DEFINITION: |f(u)-f(v)| ≤ L·|u-v| for some L
Then I have control without needing differentiability!"

AND THEN HE SAW IT:
With this assumption, the Picard iteration becomes a
CONTRACTION in the sup-norm:
||T[y] - T[z]||_∞ ≤ L·h·||y-z||_∞
If h < 1/L, then T is a strict contraction!
→ Banach's fixed point theorem gives immediate existence and uniqueness!
 

DEEPER MATHEMATICAL INSIGHT: WHY THIS WAS BRILLIANT
========================================================================

1. THE FUNDAMENTAL PROBLEM LIPSCHITZ FACED:
------------------------------------------
In the 19th century, mathematicians knew two approaches:

A. CONSTRUCTIVE (Picard, 1890):
   y_{n+1}(t) = y₀ + ∫[t₀,t] f(s, y_n(s)) ds
   Problem: How to prove {y_n} converges without strong assumptions?

B. EXISTENCE ONLY (Cauchy, 1820s, Peano 1890):
   Using Euler polygon method + compactness (Arzelà-Ascoli)
   Shows existence BUT not uniqueness!

Lipschitz wanted: CONSTRUCTIVE method that also proves UNIQUENESS.

2. THE KEY OBSERVATION:
----------------------
Differentiability gives local linear approximation:
f(u) - f(v) = f'(ξ)(u-v) for some ξ between u and v
So: |f(u)-f(v)| = |f'(ξ)|·|u-v| ≤ sup|f'|·|u-v|

But sup|f'| might not exist (f' could be unbounded)!
Example: f(y) = y² → f'(y) = 2y → unbounded as y → ∞

Lipschitz's genius: Use the INEQUALITY as the DEFINITION!

3. THE CONCEPTUAL LEAP:
----------------------
Traditional thinking: f smooth → can approximate linearly
Lipschitz's thinking: Forget smoothness! Just assume
   |f(u)-f(v)| ≤ L·|u-v| for some constant L

This is:
• Weaker than differentiability (f can be non-differentiable!)
• Stronger than mere continuity (excludes functions like √|x|)
• Quantitatively measurable (L is a concrete number)

4. WHY THIS SOLVES THE CONVERGENCE PROBLEM:
------------------------------------------
Consider the Picard operator T[y](t) = y₀ + ∫[t₀,t] f(s,y(s)) ds

For two functions y,z:
|T[y](t) - T[z](t)| ≤ ∫[t₀,t] |f(s,y(s)) - f(s,z(s))| ds
                    ≤ L·∫[t₀,t] |y(s) - z(s)| ds  ← LIPSCHITZ!
                    ≤ L·h·max_{s} |y(s)-z(s)|
                    = L·h·||y-z||_∞

Thus: ||T[y] - T[z]||_∞ ≤ L·h·||y-z||_∞

If we choose h < 1/L, then T is a CONTRACTION!
Contraction factor: q = L·h < 1

5. BANACH'S FIXED POINT THEOREM (1922, but the idea was there!):
----------------------------------------------------------------
In a complete metric space, if T is a contraction (d(Tx,Ty) ≤ q·d(x,y), q<1),
then T has a UNIQUE fixed point x* with x* = T(x*).

Applied to our case:
• Space: C([t₀-h,t₀+h]) with sup-norm (complete!)
• Operator: T as above
• Contraction: ||T[y]-T[z]|| ≤ q·||y-z|| with q = L·h < 1
→ ∃! y* such that y* = T[y*]
→ y* satisfies y*(t) = y₀ + ∫[t₀,t] f(s,y*(s)) ds
→ Differentiate: y*'(t) = f(t,y*(t)) and y*(t₀) = y₀

6. THE HISTORICAL CONTEXT - WHY THIS WAS REVOLUTIONARY:
-------------------------------------------------------
Before Lipschitz (1876):
• Cauchy (1835): Existence via Euler polygons
• Liouville (1838): Method of successive approximations
• Peano (1890): Existence without uniqueness

Lipschitz's contribution:
1. Unified construction and proof of uniqueness
2. Identified the MINIMAL sufficient condition
3. Created a template for functional analysis

7. MODERN PERSPECTIVE - THE ABSTRACT VIEW:
-----------------------------------------
Today we see Lipschitz condition as:

A. A METRIC CONDITION:
   d_Y(f(u), f(v)) ≤ L·d_X(u,v)
   where d_X, d_Y are distances in appropriate spaces

B. A GLOBAL LINEAR BOUND:
   f doesn't grow faster than linearly
   Prevents "runaway" behavior in iterations

C. A CONTRACTION CONDITION:
   Makes the Picard operator a contraction
   Enables Banach fixed point theorem

8. WHAT'S TRULY REMARKABLE:
--------------------------
The Lipschitz condition is:

• NECESSARY for uniqueness (counterexamples exist without it)
   Example: y' = √|y|, y(0)=0 has infinitely many solutions

• SUFFICIENT for both existence AND uniqueness

• PRACTICAL to verify (often from bounded derivative)

• ROBUST (preserved under many operations)

9. THE EVEN DEEPER MATHEMATICS:
-----------------------------
The Lipschitz constant L gives:

A. CONVERGENCE RATE:
   ||y_n - y*|| ≤ (L·h)^n/(1-L·h)·||y₁ - y₀||
   Geometric convergence!

B. DOMAIN OF EXISTENCE:
   h < min(1/L, b/M) where b = vertical radius, M = bound on |f|

C. SENSITIVITY TO INITIAL CONDITIONS:
   If y,z are solutions with different initial conditions:
   |y(t)-z(t)| ≤ e^{L|t-t₀|}·|y₀-z₀|
   (Continuous dependence on initial data)

10. THE LEGACY - BEYOND ODEs:
----------------------------
Lipschitz's idea spawned:

• LIPSCHITZ MANIFOLDS in differential geometry
• LIPSCHITZ FUNCTIONS in analysis (Rademacher's theorem)
• LIPSCHITZ CONTINUITY in metric spaces
• LIPSCHITZ STABILITY in control theory
• LIPSCHITZ DOMAINS in PDE theory

========================================================================
CONCLUSION: THE ESSENCE OF LIPSCHITZ'S INSIGHT
========================================================================

Lipschitz realized that to prove convergence of Picard iterations,
you don't need LOCAL linear approximation (differentiability),
but GLOBAL linear control: |f(u)-f(v)| ≤ L·|u-v|.

This single inequality:
1. Creates a recursive bound: |y_{n+1}-y_n| ≤ L·∫|y_n-y_{n-1}|
2. Yields geometric convergence when iterated
3. Makes the Picard operator a contraction
4. Guarantees both existence AND uniqueness

It was a classic case of:
"Ask not what properties the function has,
 but what properties you NEED for the proof,
 then make those properties your definition."

This is the hallmark of great mathematics:
Identifying the MINIMAL structure needed to make a proof work,
then building a theory around that structure.

_How_i_can_find_coefficients_in_presence_of_radicals_ENG_mprimes_12-12-2025.mw

vereenvoudiging_mprimes_twee_proc_4-12-2025.mw

The pattern recognition of simplify falls short for this expression and must first be converted into a recognizable form for simplify via collect.
That seems like a good explanation to me.

@salim-barzani 
determinant_-_conjugate_vraag_mprimes_29-11-2025.mw

This is working in a worksheet 

Download vraag_mprimes-_how_to_find_paramters_21-11-2021.mw

This is the secant method as a procedure for not singular elliptic curves.
I did some examples, but th eprocedure code is not yet fully functional...
Example 1 is single point and example 2 is doubling example 1 

fxy := y^2 - 2*y + 14 = 2*x^3 + 11*x^2 - 29*x - 17:
P := [3, 7];

# Test met jouw kromme
fxy := y^2 - 2*y + 14 = 2*x^3 + 11*x^2 - 29*x - 17:
P := [3, 7];

printf("=== TEST 1: POINT DOUBLING 2P ===\n");
P2 := EllipticCurveSecantMethod(fxy, x, y, P, P, -10..10, -10..15, 
                               steps = true, showplot = true);
printf("2P = (%a, %a)\n", P2[1], P2[2])

Use : m > 1 and not 3m (m= 1... n )