From a Root-Finding Problem to a Tangent Line
A root of a function \(f\) is an input \(r\) for which \(f(r)=0\). For many functions, an exact root is difficult to find directly. Newton’s method produces a sequence of estimates by replacing the function near a current estimate with its tangent line, then asking where that line crosses the \(x\)-axis.
This construction gives a specific update formula. It does not, by itself, prove that the estimates approach a root. Here the goal is to derive the formula and understand its geometric meaning, while keeping clear the conditions needed to form an update.
Deriving the Update Formula
Assume \(f\) is differentiable at \(x_n\). The tangent line to \(y=f(x)\) at the point \((x_n,f(x_n))\) has slope \(f'(x_n)\). Using the point-slope form of a line, its equation is
The \(x\)-intercept is the point on this line where \(y=0\). Write its \(x\)-coordinate as \(x_{n+1}\). Substituting \(y=0\) and \(x=x_{n+1}\) into the line equation gives
If \(f'(x_n)\ne 0\), divide by \(f'(x_n)\) and isolate \(x_{n+1}\):
The negative sign is a consequence of solving for the intercept: the function value \(f(x_n)\) is moved to the other side of the equation. The formula can also be understood as a correction to the current estimate. The correction is \(-f(x_n)/f'(x_n)\); its size and direction depend on both the height of the graph at \(x_n\) and the slope there.
Why the Tangent Line Is a Local Approximation
The tangent line is not usually the graph itself. Its role is to approximate the graph near the point where they touch. Differentiability at \(x_n\) means that, for a small change \(h\), the function can be written as
The first two terms on the right give the tangent-line prediction; \(r(h)\) is the remaining error, which becomes small relative to \(h\) as \(h\) approaches zero. Newton’s construction uses that linear prediction to estimate where the function would be zero. It does not set the full function equal to zero at the new point. The actual value \(f(x_{n+1})\) may therefore differ from zero.
This explains both the appeal and the limitation of the method. When a tangent line represents the graph well near a root, its intercept can be informative. But the construction alone gives no guarantee that the intercept is close to a root, or that repeating the construction will succeed.
Worked Tangent-Line Updates
Worked Example: A Tangent to a Quadratic
Find the Newton update from \(x_0=2\) for \(f(x)=x^2-5\).
First evaluate the function and its derivative at the current estimate:
The tangent line at \((2,-1)\) is
Set \(y=0\) to find its \(x\)-intercept:
Therefore the update is \(x_1=\frac94\). Check the result in the tangent-line equation: \(4(\frac94)-9=9-9=0\), as required for an \(x\)-intercept. The update need not be a root of \(f\): \(f(\frac94)=(\frac94)^2-5=\frac{81}{16}-\frac{80}{16}=\frac1{16}\), which is not zero.
Worked Example: A Tangent to a Square-Root Function
Find the Newton update from \(x_0=1\) for \(f(x)=\sqrt{x}-2\). The function is differentiable at \(1\), and its derivative is \(f'(x)=\frac{1}{2\sqrt{x}}\) for \(x>0\).
Evaluate the function and derivative at \(1\):
The tangent line at \((1,-1)\) is
Set the tangent-line height to zero and solve:
Thus \(x_1=3\). Substitution checks the intercept: \(\frac{3}{2}-\frac{3}{2}=0\). This verifies the tangent-line calculation; it does not assert that \(f(3)=0\).
Worked Example: A Tangent to a Reciprocal Function
For \(f(x)=\frac1x-\frac12\), find the update from \(x_0=1\). The function is defined and differentiable at \(1\), and \(f'(x)=-\frac{1}{x^2}\) for \(x\ne0\).
At the current estimate,
The tangent line at \((1,\frac12)\) is
Its \(x\)-intercept satisfies
Therefore \(x_1=\frac32\). Substituting into the line gives \(\frac32-\frac32=0\), confirming the intercept. The function itself is not zero there: \(f(\frac32)=\frac{2}{3}-\frac12=\frac{4}{6}-\frac{3}{6}=\frac16\).
Conditions and Limits of the Construction
To form a Newton update at \(x_n\), two conditions matter. First, \(f\) must be differentiable at \(x_n\), so that the tangent line is defined. Second, \(f'(x_n)\) must be nonzero, so that the tangent line has a unique \(x\)-intercept that can be found by division in the update formula.
If \(f'(x_n)=0\), the formula is undefined. Geometrically, the tangent is horizontal: if \(f(x_n)\ne0\), it does not meet the \(x\)-axis; if \(f(x_n)=0\), the current point is already a root, but the quotient in the formula still cannot be evaluated. These situations require separate consideration rather than applying the update formula as written.
There is also a domain requirement. The new estimate must be a valid input for \(f\) if you want to evaluate the function there or form another tangent. The tangent-line calculation can produce an \(x\)-coordinate outside the domain. Even when it remains in the domain, the construction alone does not establish that later estimates will approach a root.
A Common Sign Error
A frequent mistake is to move \(f(x_n)\) to the other side of the tangent-intercept equation without changing its sign. For the quadratic example, the correct equation was \(0=-1+4(x-2)\). An incorrect rearrangement would be
This answer does not satisfy the tangent-line equation: substituting \(x=\frac74\) into \(y=4x-9\) gives \(y=7-9=-2\), not \(0\). The correction is to rearrange carefully:
Checking the proposed intercept by substitution is a quick way to catch this sign error.
Geometric Meaning of the Iteration
At each stage, the graph provides a point \((x_n,f(x_n))\). The tangent line through that point is used as a simpler, linear stand-in for the graph near \(x_n\). The next estimate \(x_{n+1}\) is the point where that line crosses the horizontal axis. In this sense, the update replaces the question “Where does the graph meet the axis?” with the locally simpler question “Where does the tangent line meet the axis?”
The geometric description helps interpret the signs in the formula. If the graph is above the axis at \(x_n\), then \(f(x_n)>0\); if its tangent slope is positive, the intercept lies to the left, as the negative correction \(-f(x_n)/f'(x_n)\) indicates. If the slope is negative instead, the correction is positive. The formula records the signed horizontal movement needed to reach the tangent’s intercept.
Newton’s method is useful because it turns derivative information into a root estimate. Its essential step is the tangent-line construction, not a guarantee about what happens after that step. Understanding this distinction makes it easier to use the iteration correctly: derive the line, solve for its intercept, check the calculation, and verify that the required conditions hold at the point where the update is formed.
Check Your Understanding
Answer these questions using the tangent-line construction and its conditions.
- Starting from \(y=f(x_n)+f'(x_n)(x-x_n)\), derive the formula for \(x_{n+1}\) by setting \(y=0\).
- Why must \(f'(x_n)\ne0\) to use the Newton update formula as written?
- In the geometric interpretation, what does \(x_{n+1}\) represent?
- Why does finding the tangent line’s \(x\)-intercept not prove that \(f(x_{n+1})=0\)?
- What error is made by rearranging \(0=-1+4(x-2)\) as \(4(x-2)=-1\)?
