Free lesson · Differential calculus

Stationary points, optimisation and Newton’s method

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Start with the idea

Optimisation asks where a function is largest or smallest within the allowed domain.
A zero derivative gives a candidate interior point, not an automatic optimum.
Roots are a different problem: finding where a function itself is zero.

Symbols, units & horizon

f: objective being optimised
x*: candidate interior optimum
g: function whose zero is sought in Newton’s method
x_k: current iterate at step k
g′: derivative of the root function
k+1: next numerical iterate
J(w): utility example as a function of weight w
μ,λ,σ²: expected return, positive risk aversion and positive return variance
μ and σ² refer to the same return horizon
w: dimensionless portfolio weight, with λ scaled to the chosen utility units

When and why to use this

Use derivative signs and boundaries for hand optimisation, then use numerical root methods for equations without convenient inverses. Keep a bracketed method available when Newton is unreliable.

For a differentiable function at an unconstrained interior optimum, the slope must be zero.
Inspect curvature, endpoints, constraints and any nondifferentiable points.
A positive second derivative establishes a strict local minimum at a stationary point; a zero second derivative is inconclusive.
Newton’s method solves a root problem by replacing the function with its tangent and finding the tangent’s zero.
It is fast near some well-behaved roots but may diverge or leave the valid domain.
To optimise with Newton, apply it to the gradient, which uses curvature in the denominator.

f^{'} (x^{*}) = 0, x_{k + 1} = x_{k} - \frac{g (x_{k})}{g^{'} (x_{k})}

Calculus: derivation and arithmetic

Stationary points, optimisation and Newton’s method

For J(w)=μw−λσ²w²/2, differentiate: J′=μ−λσ²w. Set J′=0 and solve w*=μ/(λσ²). Since J″=−λσ²<0, this is the global maximum of this unconstrained quadratic.
The tangent to g at x_k is g(x_k)+g′(x_k)(x−x_k). Set it to zero and isolate x to obtain the Newton update.
To compute √2, let g(x)=x²−2 and g′=2x. Starting at 1 gives 1.5; the next step is 1.5−.25/3=1.416667.

Work it by hand

For J(w)=.02w−.1w², J′=.02−.2w gives w*=.1 and J″=−.2. On a constrained interval [0,.05], the best weight is .05 instead; the unconstrained stationary point is not feasible.

Architecture pseudocode

def quadratic_optimum(mu,risk_aversion,variance):
    if risk_aversion<=0 or variance<=0: raise ValueError("Positive inputs required")
    return mu/(risk_aversion*variance)

def newton_step(function,derivative,x):
    slope=derivative(x)
    if abs(slope)<1e-12: raise ValueError("Derivative too small")
    return x-function(x)/slope

x=1.0
for _ in range(4): x=newton_step(lambda x:x*x-2,lambda x:2*x,x)
print(x,quadratic_optimum(.02,2,.1))

An analogy to remember

A flat slope can be a hilltop, a valley, or a flat point on a continuing climb. Look around it before deciding which it is.

How this becomes a building block

First-order conditions are building blocks of portfolio optimisation. Newton-style root finding appears in yield and implied-volatility calculations, where a price error is driven toward zero. The algorithm solves the assumed equation; it does not validate the market model.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

def quadratic_optimum(mu,risk_aversion,variance):
    if risk_aversion<=0 or variance<=0: raise ValueError("Positive inputs required")
    return mu/(risk_aversion*variance)

def newton_step(function,derivative,x):
    slope=derivative(x)
    if abs(slope)<1e-12: raise ValueError("Derivative too small")
    return x-function(x)/slope

x=1.0
for _ in range(4): x=newton_step(lambda x:x*x-2,lambda x:2*x,x)
print(x,quadratic_optimum(.02,2,.1))

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