From a Polynomial to a Sequence of Estimates
Newton's method turns a starting value into a sequence of estimates for a solution of \(f(x)=0\). For a polynomial, each step requires evaluating the polynomial and its derivative at the current estimate, then finding where the tangent line meets the \(x\)-axis. The arithmetic is manageable, but the steps must be kept distinct: the tangent-line intercept is the next estimate, while the polynomial value at that estimate is its residual.
In this tutorial, exact fractions are retained as long as they are convenient. Decimal values are used to interpret the result, and substitution into the original polynomial provides a direct check. These calculations show how the method works; they do not by themselves guarantee that every starting value will lead to a root.
Why the Tangent Intercept Gives the Iteration
The tangent line to the graph of \(y=f(x)\) at \(x=x_n\) has equation \(y=f(x_n)+f'(x_n)(x-x_n)\). Its \(x\)-intercept is found by setting \(y=0\). Provided \(f'(x_n)\ne 0\), solving this linear equation gives the Newton step:
This calculation explains both the method and its condition. If the derivative is zero, the tangent is horizontal and this intercept calculation cannot be carried out by division. For each iteration, evaluate the polynomial and its derivative at the same current estimate, form the tangent line there, and then use its intercept as the next estimate.
Worked Example: Approximating a Root of \(x^2-2=0\)
Worked Example: Two Newton Steps for \(x^2-2=0\)
Let \(f(x)=x^2-2\), so \(f'(x)=2x\). Start at \(x_0=\frac32\). At this point, $$ f\left(\frac32\right)=\frac94-2=\frac14, \qquad f'\left(\frac32\right)=3. $$ The tangent line is \(y=\frac14+3(x-\frac32)\). Setting \(y=0\) gives $$ x_1=\frac32-\frac{1/4}{3} =\frac32-\frac1{12} =\frac{17}{12} \approx 1.416667. $$
Check the residual at this new estimate: $$ f\left(\frac{17}{12}\right) =\frac{289}{144}-\frac{288}{144} =\frac1{144} \approx 0.006944. $$ The derivative there is \(f'(\frac{17}{12})=\frac{17}{6}\). Thus the next tangent line is \(y=\frac1{144}+\frac{17}{6}(x-\frac{17}{12})\), and its intercept is $$ x_2=\frac{17}{12}-\frac{1/144}{17/6} =\frac{17}{12}-\frac1{408} =\frac{577}{408} \approx 1.414216. $$
Substitution verifies the new residual exactly: $$ f\left(\frac{577}{408}\right) =\frac{577^2-2(408^2)}{408^2} =\frac{332929-332928}{166464} =\frac1{166464} \approx 0.000006007. $$ As a further check on the root's location, \(f(1.4142)=1.99996164-2=-0.00003836\), whereas \(f(\frac{577}{408})=\frac1{166464}>0\). Since \(f'(x)=2x>0\) throughout this interval, \(f\) is increasing there, so the root lies between \(1.4142\) and \(1.414216\). This verifies a narrow bracket around the approximation.
Worked Example: A Root of \(x^2+x-1=0\)
Worked Example: Three Steps for \(x^2+x-1=0\)
Take \(f(x)=x^2+x-1\), with derivative \(f'(x)=2x+1\), and start at \(x_0=1\). The initial residual and derivative are $$ f(1)=1, \qquad f'(1)=3. $$ The tangent line is \(y=1+3(x-1)\), so its intercept gives $$ x_1=1-\frac13=\frac23\approx 0.666667. $$
At \(x_1=\frac23\), the residual and derivative are $$ f\left(\frac23\right)=\frac49+\frac23-1=\frac19, \qquad f'\left(\frac23\right)=\frac73. $$ The next tangent line is \(y=\frac19+\frac73(x-\frac23)\). Its intercept is $$ x_2=\frac23-\frac{1/9}{7/3} =\frac23-\frac1{21} =\frac{13}{21} \approx 0.619048. $$
Substitute \(x_2\) into the polynomial: $$ f\left(\frac{13}{21}\right) =\frac{169}{441}+\frac{273}{441}-\frac{441}{441} =\frac1{441}. $$ Also \(f'(\frac{13}{21})=\frac{47}{21}\), so the next tangent line is \(y=\frac1{441}+\frac{47}{21}(x-\frac{13}{21})\). Therefore $$ x_3=\frac{13}{21}-\frac{1/441}{47/21} =\frac{13}{21}-\frac1{987} =\frac{610}{987} \approx 0.618034. $$
Finally, check this estimate by substitution: $$ f\left(\frac{610}{987}\right) =\frac{610^2+610(987)-987^2}{987^2} =\frac{372100+602070-974169}{974169} =\frac1{974169} \approx 0.000001027. $$ For a sign check, \(f(0.618)=0.618^2+0.618-1=-0.000076\), while the exact residual at \(x_3\) is positive. Since \(f'(x)=2x+1>0\) for positive \(x\), the root lies between \(0.618\) and \(0.618034\ldots\). The substitution and sign check agree with the computed estimate.
Worked Example: A Cubic with a Negative Tangent Slope
Worked Example: Two Steps for \(x^3-3x+1=0\)
Let \(f(x)=x^3-3x+1\), so \(f'(x)=3x^2-3\), and use \(x_0=0\). The residual is \(f(0)=1\), and the derivative is \(f'(0)=-3\). The tangent line is \(y=1-3(x-0)\), whose intercept is $$ x_1=0-\frac{1}{-3}=\frac13\approx 0.333333. $$ The negative slope makes the correction positive.
At \(x_1=\frac13\), compute both values: $$ f\left(\frac13\right) =\frac1{27}-1+1 =\frac1{27}, \qquad f'\left(\frac13\right)=\frac13-3=-\frac83. $$ The tangent line is \(y=\frac1{27}-\frac83(x-\frac13)\). Its intercept gives $$ x_2=\frac13-\frac{1/27}{-8/3} =\frac13+\frac1{72} =\frac{25}{72} \approx 0.347222. $$
Substitution checks the residual: $$ f\left(\frac{25}{72}\right) =\frac{15625}{373248}-\frac{75}{72}+1 =\frac{15625-388800+373248}{373248} =\frac{73}{373248} \approx 0.00019558. $$ The residual is small, though not zero; \(x_2\) is an approximation, not an exact solution. The substitution confirms how closely it satisfies the equation at this stage.
Reading Residuals and Checking the Arithmetic
At each step, keep three quantities conceptually separate: the current estimate \(x_n\), the residual \(f(x_n)\), and the derivative \(f'(x_n)\). The residual and derivative determine the correction. They are not themselves the next estimate. A useful written record is:
| Quantity | What it tells you |
|---|---|
| \(x_n\) | The current input to the polynomial |
| \(f(x_n)\) | The residual at that input |
| \(f'(x_n)\) | The slope of the tangent line at that input |
| \(x_{n+1}\) | The tangent line's \(x\)-intercept |
A small residual is a useful substitution check: it says the polynomial value at the displayed estimate is close to zero. But residual size alone is not a universal measure of distance to a root. Polynomial values can change slowly or rapidly with \(x\), depending on the location and slope. When possible, a sign change on an interval, together with monotonicity there, gives a stronger check on where a root lies. In the first two examples, those checks were possible because the derivatives have a fixed positive sign on the intervals considered.
Precision also matters. Rounding an iterate early and then using it in later calculations can change subsequent values. For hand calculations, retain fractions when practical; for decimal calculations, keep extra digits during intermediate steps and state the precision of the final approximation. Always evaluate \(f\) and \(f'\) at the same iterate before computing its tangent intercept.
A Common Mistake: Omitting the Derivative
For \(f(x)=x^2-2\) at \(x_0=\frac32\), a tempting but incorrect calculation is $$ x_1=\frac32-\frac14=\frac54. $$ This subtracts the residual directly. It ignores the tangent slope \(f'(\frac32)=3\), so it is not the tangent-line intercept. The correct correction is the residual divided by the derivative: $$ x_1=\frac32-\frac{f(3/2)}{f'(3/2)} =\frac32-\frac{1/4}{3} =\frac32-\frac1{12} =\frac{17}{12}. $$
The derivative must also be evaluated at the current iterate, not at the new one. The reliable routine is: calculate \(f(x_n)\), calculate \(f'(x_n)\), form \(x_{n+1}=x_n-f(x_n)/f'(x_n)\), and then substitute \(x_{n+1}\) into \(f\) to check its residual.
Check Your Understanding
Try each question by writing down the polynomial value, derivative, and tangent intercept explicitly.
- For \(f(x)=x^2-3\) and \(x_0=2\), find the tangent line and calculate \(x_1\).
- For \(f(x)=x^2-3\) at the estimate from Question 1, calculate the residual exactly.
- For \(f(x)=x^2+x-1\) and \(x_0=0\), calculate one Newton iterate and its residual.
- What condition on \(f'(x_n)\) is required to compute the Newton step by the iteration formula?
- Why does a small residual provide a substitution check but not, by itself, a universal bound on the distance to a root?
