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Newton’s Method

Describe the steps of Newton’s method.

Describing Newton’s Method

Consider the task of finding the solutions of \(f(x)=0.\) If \(f\) is the first-degree polynomial \(f(x)=ax+b,\) then the solution of \(f(x)=0\) is given by the formula \(x=-\frac{b}{a}.\) If \(f\) is the second-degree polynomial \(f(x)=a{x}^{2}+bx+c,\) the solutions of \(f(x)=0\) can be found by using the quadratic formula. However, for polynomials of degree \(3\) or more, finding roots of \(f\) becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if \(f\) is a polynomial of degree \(5\) or greater, it is known that no such formulas exist. For example, consider the function

\[f(x)={x}^{5}+8{x}^{4}+4{x}^{3}-2x-7.\]

No formula exists that allows us to find the solutions of \(f(x)=0.\) Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of \(\text{tan}(x)-x=0.\) No simple formula exists for the solutions of this equation. In cases such as these, we can use Newton’s method to approximate the roots.

Newton’s method makes use of the following idea to approximate the solutions of \(f(x)=0.\) By sketching a graph of \(f,\) we can estimate a root of \(f(x)=0.\) Let’s call this estimate \({x}_{0}.\) We then draw the tangent line to \(f\) at \({x}_{0}.\) If \({f}^{'}({x}_{0})\ne 0,\) this tangent line intersects the \(x\)-axis at some point \(({x}_{1},0).\) Now let \({x}_{1}\) be the next approximation to the actual root. Typically, \({x}_{1}\) is closer than \({x}_{0}\) to an actual root. Next we draw the tangent line to \(f\) at \({x}_{1}.\) If \({f}^{'}({x}_{1})\ne 0,\) this tangent line also intersects the \(x\)-axis, producing another approximation, \({x}_{2}.\) We continue in this way, deriving a list of approximations: \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }.\) Typically, the numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) quickly approach an actual root \(x*,\) as shown in the following figure.

Now let’s look at how to calculate the approximations \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }.\) If \({x}_{0}\) is our first approximation, the approximation \({x}_{1}\) is defined by letting \(({x}_{1},0)\) be the \(x\)-intercept of the tangent line to \(f\) at \({x}_{0}.\) The equation of this tangent line is given by

\[y=f({x}_{0})+{f}^{'}({x}_{0})(x-{x}_{0}).\]

Therefore, \({x}_{1}\) must satisfy

\[f({x}_{0})+{f}^{'}({x}_{0})({x}_{1}-{x}_{0})=0.\]

Solving this equation for \({x}_{1},\) we conclude that

\[{x}_{1}={x}_{0}-\frac{f({x}_{0})}{f'({x}_{0})}.\]

Similarly, the point \(({x}_{2},0)\) is the \(x\)-intercept of the tangent line to \(f\) at \({x}_{1}.\) Therefore, \({x}_{2}\) satisfies the equation

\[{x}_{2}={x}_{1}-\frac{f({x}_{1})}{f'({x}_{1})}.\]\[{x}_{n}={x}_{n-1}-\frac{f({x}_{n-1})}{f'({x}_{n-1})}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Failures of Newton’s Method

Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s method might fail include the following:

  1. At one of the approximations \({x}_{n},\) the derivative \({f}^{'}\) is zero at \({x}_{n},\) but \(f({x}_{n})\ne 0.\) As a result, the tangent line of \(f\) at \({x}_{n}\) does not intersect the \(x\)-axis. Therefore, we cannot continue the iterative process.
  2. The approximations \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) may approach a different root. If the function \(f\) has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see ). This event most often occurs when we do not choose the approximation \({x}_{0}\) close enough to the desired root.
  3. The approximations may fail to approach a root entirely. In , we provide an example of a function and an initial guess \({x}_{0}\) such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values.
Example

Try it.

Consider the function \(f(x)={x}^{3}-2x+2.\) Let \({x}_{0}=0.\) Show that the sequence \({x}_{1},{x}_{2}\text{,\ldots }\) fails to approach a root of \(f.\)

Solution

For \(f(x)={x}^{3}-2x+2,\) the derivative is \({f}^{'}(x)=3{x}^{2}-2.\) Therefore,

\[{x}_{1}={x}_{0}-\frac{f({x}_{0})}{{f}^{'}({x}_{0})}=0-\frac{f(0)}{{f}^{'}(0)}=-\frac{2}{-2}=1.\]

In the next step,

\[{x}_{2}={x}_{1}-\frac{f({x}_{1})}{f'({x}_{1})}=1-\frac{f(1)}{{f}^{'}(1)}=1-\frac{1}{1}=0.\]

Consequently, the numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) continue to bounce back and forth between \(0\) and \(1\) and never get closer to the root of \(f\) which is over the interval \([-2,-1]\) (see ). Fortunately, if we choose an initial approximation \({x}_{0}\) closer to the actual root, we can avoid this situation.

From , we see that Newton’s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton’s method are included in texts on numerical analysis.

Other Iterative Processes

As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of iterative process.

Consider a function \(F\) and an initial number \({x}_{0}.\) Define the subsequent numbers \({x}_{n}\) by the formula \({x}_{n}=F({x}_{n-1}).\) This process is an iterative process that creates a list of numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots },{x}_{n}\text{,\ldots }.\) This list of numbers may approach a finite number \(x*\) as \(n\) gets larger, or it may not. In , we see an example of a function \(F\) and an initial guess \({x}_{0}\) such that the resulting list of numbers approaches a finite value.

Example

Try it.

Let \(F(x)=\frac{1}{2}x+4\) and let \({x}_{0}=0.\) For all \(n\ge 1,\) let \({x}_{n}=F({x}_{n-1}).\) Find the values \({x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}.\) Make a conjecture about what happens to this list of numbers \({x}_{1},{x}_{2},{x}_{3}\text{\ldots },{x}_{n}\text{,\ldots }\) as \(n\to \infty .\) If the list of numbers \({x}_{1},{x}_{2},{x}_{3}\text{,\ldots }\) approaches a finite number \(x*,\) then \(x*\) satisfies \(x*=F(x*),\) and \(x*\) is called a fixed point of \(F.\)

Solution

If \({x}_{0}=0,\) then

\[\begin{array}{l} \\ \\ {x}_{1}=\frac{1}{2}(0)+4=4 \\ {x}_{2}=\frac{1}{2}(4)+4=6 \\ {x}_{3}=\frac{1}{2}(6)+4=7 \\ {x}_{4}=\frac{1}{2}(7)+4=7.5 \\ {x}_{5}=\frac{1}{2}(7.5)+4=7.75 \\ {x}_{6}=\frac{1}{2}(7.75)+4=7.875 \\ {x}_{7}=\frac{1}{2}(7.875)+4=7.9375 \\ {x}_{8}=\frac{1}{2}(7.9375)+4=7.96875 \\ {x}_{9}=\frac{1}{2}(7.96875)+4=7.984375.\end{array}\]

From this list, we conjecture that the values \({x}_{n}\) approach \(8.\)

provides a graphical argument that the values approach \(8\) as \(n\to \infty .\) Starting at the point \(({x}_{0},{x}_{0}),\) we draw a vertical line to the point \(({x}_{0},F({x}_{0})).\) The next number in our list is \({x}_{1}=F({x}_{0}).\) We use \({x}_{1}\) to calculate \({x}_{2}.\) Therefore, we draw a horizontal line connecting \(({x}_{0},{x}_{1})\) to the point \(({x}_{1},{x}_{1})\) on the line \(y=x,\) and then draw a vertical line connecting \(({x}_{1},{x}_{1})\) to the point \(({x}_{1},F({x}_{1})).\) The output \(F({x}_{1})\) becomes \({x}_{2}.\) Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines \(F(x)=\frac{x}{2}+4\) and \(y=x.\) The line segments get closer to the intersection point of these two lines, which occurs when \(x=F(x).\) Solving the equation \(x=\frac{x}{2}+4,\) we conclude they intersect at \(x=8.\) Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) approaches \(x*=8\) as \(n\to \infty .\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

  • Newton’s method approximates roots of \(f(x)=0\) by starting with an initial approximation \({x}_{0},\) then uses tangent lines to the graph of \(f\) to create a sequence of approximations \({x}_{1},{x}_{2},{x}_{3}\text{,\ldots }.\)
  • Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method fails to work because the list of numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) does not approach a finite value or it approaches a value other than the root sought.
  • Any process in which a list of numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) is generated by defining an initial number \({x}_{0}\) and defining the subsequent numbers by the equation \({x}_{n}=F({x}_{n-1})\) for some function \(F\) is an iterative process. Newton’s method is an example of an iterative process, where the function \(F(x)=x-[\frac{f(x)}{{f}^{'}(x)}]\) for a given function \(f.\)

Newton’s Method

For the following exercises, write Newton’s formula as \({x}_{n+1}=F({x}_{n})\) for solving \(f(x)=0.\)

For the following exercises, solve \(f(x)=0\) using the iteration \({x}_{n+1}={x}_{n}-cf({x}_{n}),\) which differs slightly from Newton’s method. Find a \(c\) that works and a \(c\) that fails to converge, with the exception of \(c=0.\)

For the following exercises, start at

a. \({x}_{0}=0.6\) and

b. \({x}_{0}=2.\)

Compute \({x}_{1}\) and \({x}_{2}\) using the specified iterative method.

For the following exercises, solve to four decimal places using Newton’s method and a computer or calculator. Choose any initial guess \({x}_{0}\) that is not the exact root.

\[{x}_{n}={x}_{n-1}-f({x}_{n-1})\frac{{x}_{n-1}-{x}_{n-2}}{f({x}_{n-1})-f({x}_{n-2})}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Use Newton’s method to approximate a root of \(f(x)={x}^{3}-3x+1\) in the interval \([1,2].\) Let \({x}_{0}=2\) and find \({x}_{1},{x}_{2},{x}_{3},{x}_{4},\) and \({x}_{5}.\)

    Révèle la réponse

    From , we see that \(f\) has one root over the interval \((1,2).\) Therefore \({x}_{0}=2\) seems like a reasonable first approximation. To find the next approximation, we use . Since \(f(x)={x}^{3}-3x+1,\) the derivative is \({f}^{'}(x)=3{x}^{2}-3.\) Using with \(n=1\) (and a calculator that displays \(10\) digits), we obtain

    \[{x}_{1}={x}_{0}-\frac{f({x}_{0})}{f'({x}_{0})}=2-\frac{f(2)}{f'(2)}=2-\frac{3}{9}\approx 1.666666667.\]

    To find the next approximation, \({x}_{2},\) we use with \(n=2\) and the value of \({x}_{1}\) stored on the calculator. We find that

    \[{x}_{2}={x}_{1}-\frac{f({x}_{1})}{f'({x}_{1})}\approx 1.548611111.\]

    Continuing in this way, we obtain the following results:

    \[\begin{array}{l}{x}_{1}\approx 1.666666667 \\ {x}_{2}\approx 1.548611111 \\ {x}_{3}\approx 1.532390162 \\ {x}_{4}\approx 1.532088989 \\ {x}_{5}\approx 1.532088886 \\ {x}_{6}\approx 1.532088886.\end{array}\]

    We note that we obtained the same value for \({x}_{5}\) and \({x}_{6}.\) Therefore, any subsequent application of Newton’s method will most likely give the same value for \({x}_{n}.\)

  2. Letting \({x}_{0}=0,\) let’s use Newton’s method to approximate the root of \(f(x)={x}^{3}-3x+1\) over the interval \([0,1]\) by calculating \({x}_{1}\) and \({x}_{2}.\)

    Révèle la réponse

    \({x}_{1}\approx 0.33333333,{x}_{2}\approx 0.347222222\)

  3. Use Newton’s method to approximate \(\sqrt{2}\) (). Let \(f(x)={x}^{2}-2,\) let \({x}_{0}=2,\) and calculate \({x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}.\) (We note that since \(f(x)={x}^{2}-2\) has a zero at \(\sqrt{2},\) the initial value \({x}_{0}=2\) is a reasonable choice to approximate \(\sqrt{2}.)\)

    Révèle la réponse

    For \(f(x)={x}^{2}-2,{f}^{'}(x)=2x.\) From , we know that

    \[\begin{array}{ll}{x}_{n} & ={x}_{n-1}-\frac{f({x}_{n-1})}{f'({x}_{n-1})} \\ & ={x}_{n-1}-\frac{{x}^{2}{}_{n-1}-2}{2{x}_{n-1}} \\ & =\frac{1}{2}{x}_{n-1}+\frac{1}{{x}_{n-1}} \\ & =\frac{1}{2}({x}_{n-1}+\frac{2}{{x}_{n-1}}).\end{array}\]

    Therefore,

    \[\begin{array}{l} \\ \\ {x}_{1}=\frac{1}{2}({x}_{0}+\frac{2}{{x}_{0}})=\frac{1}{2}(2+\frac{2}{2})=1.5 \\ {x}_{2}=\frac{1}{2}({x}_{1}+\frac{2}{{x}_{1}})=\frac{1}{2}(1.5+\frac{2}{1.5})\approx 1.416666667.\end{array}\]

    Continuing in this way, we find that

    \[\begin{array}{l}{x}_{1}=1.5 \\ {x}_{2}\approx 1.416666667 \\ {x}_{3}\approx 1.414215686 \\ {x}_{4}\approx 1.414213562 \\ {x}_{5}\approx 1.414213562.\end{array}\]

    Since we obtained the same value for \({x}_{4}\) and \({x}_{5},\) it is unlikely that the value \({x}_{n}\) will change on any subsequent application of Newton’s method. We conclude that \(\sqrt{2}\approx 1.414213562.\)

  4. Use Newton’s method to approximate \(\sqrt{3}\) by letting \(f(x)={x}^{2}-3\) and \({x}_{0}=3.\) Find \({x}_{1}\) and \({x}_{2}.\)

    Révèle la réponse

    \({x}_{1}=2,{x}_{2}=1.75\)

  5. Consider the function \(f(x)={x}^{3}-2x+2.\) Let \({x}_{0}=0.\) Show that the sequence \({x}_{1},{x}_{2}\text{,\ldots }\) fails to approach a root of \(f.\)

    Révèle la réponse

    For \(f(x)={x}^{3}-2x+2,\) the derivative is \({f}^{'}(x)=3{x}^{2}-2.\) Therefore,

    \[{x}_{1}={x}_{0}-\frac{f({x}_{0})}{{f}^{'}({x}_{0})}=0-\frac{f(0)}{{f}^{'}(0)}=-\frac{2}{-2}=1.\]

    In the next step,

    \[{x}_{2}={x}_{1}-\frac{f({x}_{1})}{f'({x}_{1})}=1-\frac{f(1)}{{f}^{'}(1)}=1-\frac{1}{1}=0.\]

    Consequently, the numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) continue to bounce back and forth between \(0\) and \(1\) and never get closer to the root of \(f\) which is over the interval \([-2,-1]\) (see ). Fortunately, if we choose an initial approximation \({x}_{0}\) closer to the actual root, we can avoid this situation.

  6. For \(f(x)={x}^{3}-2x+2,\) let \({x}_{0}=-1.5\) and find \({x}_{1}\) and \({x}_{2}.\)

    Révèle la réponse

    \({x}_{1}\approx -1.842105263,{x}_{2}\approx -1.772826920\)

  7. Let \(F(x)=\frac{1}{2}x+4\) and let \({x}_{0}=0.\) For all \(n\ge 1,\) let \({x}_{n}=F({x}_{n-1}).\) Find the values \({x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}.\) Make a conjecture about what happens to this list of numbers \({x}_{1},{x}_{2},{x}_{3}\text{\ldots },{x}_{n}\text{,\ldots }\) as \(n\to \infty .\) If the list of numbers \({x}_{1},{x}_{2},{x}_{3}\text{,\ldots }\) approaches a finite number \(x*,\) then \(x*\) satisfies \(x*=F(x*),\) and \(x*\) is called a fixed point of \(F.\)

    Révèle la réponse

    If \({x}_{0}=0,\) then

    \[\begin{array}{l} \\ \\ {x}_{1}=\frac{1}{2}(0)+4=4 \\ {x}_{2}=\frac{1}{2}(4)+4=6 \\ {x}_{3}=\frac{1}{2}(6)+4=7 \\ {x}_{4}=\frac{1}{2}(7)+4=7.5 \\ {x}_{5}=\frac{1}{2}(7.5)+4=7.75 \\ {x}_{6}=\frac{1}{2}(7.75)+4=7.875 \\ {x}_{7}=\frac{1}{2}(7.875)+4=7.9375 \\ {x}_{8}=\frac{1}{2}(7.9375)+4=7.96875 \\ {x}_{9}=\frac{1}{2}(7.96875)+4=7.984375.\end{array}\]

    From this list, we conjecture that the values \({x}_{n}\) approach \(8.\)

    provides a graphical argument that the values approach \(8\) as \(n\to \infty .\) Starting at the point \(({x}_{0},{x}_{0}),\) we draw a vertical line to the point \(({x}_{0},F({x}_{0})).\) The next number in our list is \({x}_{1}=F({x}_{0}).\) We use \({x}_{1}\) to calculate \({x}_{2}.\) Therefore, we draw a horizontal line connecting \(({x}_{0},{x}_{1})\) to the point \(({x}_{1},{x}_{1})\) on the line \(y=x,\) and then draw a vertical line connecting \(({x}_{1},{x}_{1})\) to the point \(({x}_{1},F({x}_{1})).\) The output \(F({x}_{1})\) becomes \({x}_{2}.\) Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines \(F(x)=\frac{x}{2}+4\) and \(y=x.\) The line segments get closer to the intersection point of these two lines, which occurs when \(x=F(x).\) Solving the equation \(x=\frac{x}{2}+4,\) we conclude they intersect at \(x=8.\) Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers \({x}_{0},{x}_{1},{x}_{2}\text{,\ldots }\) approaches \(x*=8\) as \(n\to \infty .\)

  8. Consider the function \(F(x)=\frac{1}{3}x+6.\) Let \({x}_{0}=0\) and let \({x}_{n}=F({x}_{n-1})\) for \(n\ge 2.\) Find \({x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}.\) Make a conjecture about what happens to the list of numbers \({x}_{1},{x}_{2},{x}_{3}\text{,\ldots }{x}_{n}\text{,\ldots }\) as \(n\to \infty .\)

    Révèle la réponse

    \({x}_{1}=6,{x}_{2}=8,{x}_{3}=\frac{26}{3},{x}_{4}=\frac{80}{9},{x}_{5}=\frac{242}{27};x*=9\)

  9. \(f(x)={x}^{2}+1\)

  10. \(f(x)={x}^{3}+2x+1\)

    Révèle la réponse

    \(F({x}_{n})={x}_{n}-\frac{{x}_{n}{}^{3}+2{x}_{n}+1}{3{x}_{n}{}^{2}+2}\)

  11. \(f(x)=\text{sin}\ x\)

  12. \(f(x)={e}^{x}\)

    Révèle la réponse

    \(F({x}_{n})={x}_{n}-\frac{{e}^{{x}_{n}}}{{e}^{{x}_{n}}}\)

  13. \(f(x)={x}^{3}+3x{e}^{x}\)

  14. \(f(x)={x}^{2}-4,\) with \({x}_{0}=0\)

    Révèle la réponse

    \(|c|>0.5\) fails, \(|c|\le 0.5\) works

  15. \(f(x)={x}^{2}-4x+3,\) with \({x}_{0}=2\)

  16. What is the value of \(\text{“}c\text{”}\) for Newton’s method?

    Révèle la réponse

    \(c=\frac{1}{{f}^{'}({x}_{n})}\)

  17. \({x}_{n+1}={x}_{n}{}^{2}-\frac{1}{2}\)

  18. \({x}_{n+1}=2{x}_{n}(1-{x}_{n})\)

    Révèle la réponse

    a. \({x}_{1}=\frac{12}{25},{x}_{2}=\frac{312}{625};\) b. \({x}_{1}=-4,{x}_{2}=-40\)

  19. \({x}_{n+1}=\sqrt{{x}_{n}}\)

  20. \({x}_{n+1}=\frac{1}{\sqrt{{x}_{n}}}\)

    Révèle la réponse

    a. \({x}_{1}=1.291,{x}_{2}=0.8801;\) b. \({x}_{1}=0.7071,{x}_{2}=1.189\)

  21. \({x}_{n+1}=3{x}_{n}(1-{x}_{n})\)

  22. \({x}_{n+1}={x}_{n}{}^{2}+{x}_{n}-2\)

    Révèle la réponse

    a. \({x}_{1}=-\frac{26}{25},{x}_{2}=-\frac{1224}{625};\) b. \({x}_{1}=4,{x}_{2}=18\)

  23. \({x}_{n+1}=\frac{1}{2}{x}_{n}-1\)

  24. \({x}_{n+1}=|{x}_{n}|\)

    Révèle la réponse

    a. \({x}_{1}=\frac{6}{10},{x}_{2}=\frac{6}{10};\) b. \({x}_{1}=2,{x}_{2}=2\)

  25. \({x}^{2}-10=0\)

  26. \({x}^{4}-100=0\)

    Révèle la réponse

    \(3.1623\ \text{or}\ -3.1623\)

  27. \({x}^{2}-x=0\)

  28. \({x}^{3}-x=0\)

    Révèle la réponse

    \(0,-1\ \text{or}\ 1\)

  29. \(x+5\ \text{cos}(x)=0\)

  30. \(x+\text{tan}(x)=0,\) choose \({x}_{0}\in (-\frac{\pi }{2},\frac{\pi }{2})\)

    Révèle la réponse

    \(0\)

  31. \(\frac{1}{1-x}=2\)

  32. \(1+x+{x}^{2}+{x}^{3}+{x}^{4}=2\)

    Révèle la réponse

    \(0.5188\ \text{or}\ -1.2906\)

  33. \({x}^{3}+{(x+1)}^{3}={10}^{3}\)

  34. \(x={\text{sin}}^{2}(x)\)

    Révèle la réponse

    \(0\)

  35. \(\text{sin}\ x\)

  36. \(\text{tan}(x)\) on \(x=(\frac{\pi }{2},\frac{3\pi }{2})\)

    Révèle la réponse

    \(4.493\)

  37. \({e}^{x}-2\)

  38. \(\text{ln}(x)+2\)

    Révèle la réponse

    \(0.159,3.146\)

  39. To find candidates for maxima and minima, we need to find the critical points \({f}^{'}(x)=0.\) Show that to solve for the critical points of a function \(f(x),\) Newton’s method is given by \({x}_{n+1}={x}_{n}-\frac{{f}^{'}({x}_{n})}{{f}^{″}({x}_{n})}.\)

  40. What additional restrictions are necessary on the function \(f?\)

    Révèle la réponse

    We need \(f\) to be twice continuously differentiable.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\approx
approximately equal
Equal to the precision shown, not exactly.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\neq
not equal
The two sides are different.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.

How to: Newton’s Method

  1. Describe the steps of Newton’s method.
  2. Explain what an iterative process means.
  3. Recognize when Newton’s method does not work.
  4. Apply iterative processes to various situations.
  5. At one of the approximations
  6. The approximations
  7. The approximations may fail to approach a root entirely. In
  8. Let

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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