One of the most used tools in mathematics are functions, since they are the ones that allow you to model the behavior of many things. Most of the time, the functions that are used are very complex and that is why people usually work with simpler approximations.
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Sign up for freeThe first degree Taylor polynomial is also called ____.
True or False: Taylor polynomials allow you to approximate any function you can take derivatives of using powers of \(x\).
The third degree Taylor polynomial of a function is also called a ____.
The second degree Taylor polynomial of a function is also called a ____.
When calculating Taylor polynomials:
If you have the Taylor polynomial of a function \(f\) centered at \(x=2\). What values of \(f\) can you approximate decently using this polynomial?
If you want to calculate the fifth degree Taylor polynomial of a function, how many derivatives do you have to calculate?
Is it safe to estimate the value of \(f(49)\) using a Taylor polynomial centered at \(x=50\)?
If you want to estimate the value of \(\sqrt{90}\) using the function \(\sqrt{x}\), where is it best to center your Taylor polynomial?
The first degree Taylor polynomial is also called ____.
True or False: Taylor polynomials allow you to approximate any function you can take derivatives of using powers of \(x\).
The third degree Taylor polynomial of a function is also called a ____.
The second degree Taylor polynomial of a function is also called a ____.
When calculating Taylor polynomials:
If you have the Taylor polynomial of a function \(f\) centered at \(x=2\). What values of \(f\) can you approximate decently using this polynomial?
If you want to calculate the fifth degree Taylor polynomial of a function, how many derivatives do you have to calculate?
Is it safe to estimate the value of \(f(49)\) using a Taylor polynomial centered at \(x=50\)?
If you want to estimate the value of \(\sqrt{90}\) using the function \(\sqrt{x}\), where is it best to center your Taylor polynomial?
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Team Taylor Polynomials Teachers
There are many types of approximations, among them are Taylor polynomials. In this article, you will learn what a Taylor polynomial is, how it is calculated and how it can be used.
Sometimes to avoid working with very complex functions, people use approximations. The most commonly used approximation is the linear or first order approximation (see the article linear approximations and differentials, for more information about this).
Given a function \(f(x)\), the linear approximation of \(f(x)\) at the point \(x=c\) is given by the function
\[L(x)=f(c)+f'(c)(x-c).\]
Let's look at a quick example.
For instance, let \(f(x)=\ln(x)\), then the linear approximation of \(f(x)\) at the point \(x=1\) is given by
\[\begin{align} L(x)&=\ln (1)+\frac{1}{1}(x-1)\\ &=x-1. \end{align}\]
Figure 1. Graph of the function \(\ln(x)\) with its linear approximation.
So, how can you get a better approximation that works for values farther away?
Well, you can think of the following: the linear approximation works because you know the rate of change at a point, if you knew how the rate of change varies (i.e., the second derivative), that might provide you more information about what the function is like.
In general, if you knew how “all” the derivatives of the function behave, you could know exactly what the function is like. That is the idea behind Taylor polynomials.
Let's state the definition of the Taylor polynomial.
Let \(f\) be a function with at least \(n\) derivatives at \(x=c\). Then, the \(n^{th}\) order Taylor polynomial centered at \(x=c\) is given by
\[\begin{align} T_n(x)&=f(c)+\frac{f'(c)(x-c)}{1!}+\frac{f''(c)(x-c)^2}{2!}+\dots\\ & \quad +\frac{f^{(n)}(c)(x-c)^n}{n!}.\end{align}\]
This polynomial of degree \(n\) has the property that
\[T_n^{(k)}(c)=f^{(k)}(c)\]
for \(k=0,\dots,n\), and approximates \(f(x)\) near \(x=c\).
Note that the first degree Taylor polynomial is the same as the linear approximation (also called the tangent line)!
There is a special case when \(x=0\) because it is much easier to write.
Let \(f\) be a function with at least \(n\) derivatives at \(x=0\). Then, the \(n^{th}\) order Maclaurin polynomial centered at \(x=0\) is given by
\[\begin{align}M_n(x)&=f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+\dots\\ & \quad +\frac{f^{(n)}(0)x^n}{n!}.\end{align}\]
It is worth mentioning that the Taylor polynomials allow you to approximate any function using powers of \(x\). How good an approximation is it? That will depend on how many derivatives you calculate, the more derivatives, the more accurate the approximation.
To learn how to calculate the error in the approximation, visit our article called Lagrange Error Bound.
A particular case is when you want to approximate a function using the second derivative.
Let \(f\) be a function with at least \(2\) derivatives at \(x=c\). The second degree Taylor polynomial, or quadratic approximation, centered at \(x=c\) is given by the function\[T_2(x)=f(c)+\frac{f'(c)(x-c)}{1!}+\frac{f''(c)(x-c)^{2}}{2!}.\]
To see how a quadratic approximation is better than a linear approximation, let's approximate the function \(f(x)=\sin x\) at \(x=\dfrac{\pi}{2}\). First, let's calculate the derivatives and evaluate the functions at \(x=\dfrac{\pi}{2}\).
Table 1. Function and derivative values for \(\sin x\) at \(\dfrac{\pi}{2}\).
\(f(x)=\sin x\) | \(f\left( \dfrac{\pi}{2}\right)=1\) |
\(f'(x)=\cos x\) | \(f'\left( \dfrac{\pi}{2}\right) =0\) |
\(f''(x)=-\sin x\) | \(f''\left( \dfrac{\pi}{2}\right) =-1\) |
Thus, the first degree Taylor polynomial for \(f(x)=\sin x\) at \(x=\dfrac{\pi}{2}\) is
\[T_1(x)=1,\]
and the second degree Taylor polynomial for \(f(x)=\sin x\) at \(x=\dfrac{\pi}{2}\) is
\[T_2(x)=1-\frac{(x-\frac{\pi}{2})^2}{2!}.\]
Figure 2. Graph of the function \(f(x)=\sin x\) with its first and second degree Taylor polynomial.
Note that approximations also allow you to estimate values of a function at points where it is difficult to evaluate. For example, you know that \(\sin(\frac{\pi}{2})\) is equal to \(1\), but what about the value of \(\sin 2\)? Using the quadratic approximation, you get
\[\begin{align} \sin 2 &\approx T_2(2) \\ &=1-\frac{(2-\frac{\pi}{2})^2}{2!} \\ &\approx 0.9. \end{align}\]
❗❗ Remember that a Taylor polynomial centered at \(x=a\), only allows you to estimate values near \(x=a\).
You could also approximate a function using the third derivative.
Let \(f\) be a function with at least \(3\) derivatives at \(x=c\). The third degree Taylor polynomial, or cubic approximation, centered at \(x=c\) is \[\begin{align} T_3(x)&=f(c)+\frac{f'(c)(x-c)}{1!}+\frac{f''(c)(x-c)^{2}}{2!}\\ &\quad +\frac{f'''(c)(x-c)^{3}}{3!}.\end{align}\]
Let's calculate the third degree Taylor polynomial for the function \(g(x)=\sqrt{x}\) at \(x=1\).
Table 2. Derivatives and function values for \(g(x) = \sqrt{x}\).
\(g(x)=\sqrt{x}\) | \(g(1)=1\) |
\(g(x)=\dfrac{1}{2\sqrt{x}}\) | \(g(1)=\dfrac{1}{2}\) |
\(g(x)=-\dfrac{1}{4x^{3/2}}\) | \(g(1)=-\dfrac{1}{4}\) |
\(g(x)=\dfrac{3}{8x^{5/2}}\) | \(g(1)=\dfrac{3}{8}\) |
Thus, its third degree Taylor polynomial centered at \(x=1\) is
\[T_3(x)=1+\frac{(x-1)}{2}-\frac{(x-1)^2}{8}+\frac{3(x-1)^3}{48}.\]
Figure 3. Graph of the function \(g(x)=\sqrt{x}\) with its Taylor polynomial of third degree.
Let's look at more examples of how to use Taylor polynomials to estimate values of a function.
Calculate the fourth degree Taylor polynomial for the function \(f(x)=\cos x\) at \(x=0\) and use it to approximate \(\cos\left(\dfrac{\pi}{2}\right)\).
Solution:
Remember that a Taylor polynomial at \(x=0\) is called a Maclaurin polynomial! First, calculate the first \(4\) derivatives of \(f(x)=\cos x\) and evaluate them at \(x=0\).
Table 3. Derivatives and function values for \(\cos x\).
| \(f(x)=\cos{x}\) | \(f(0)=1\) |
| \(f'(x)=-\sin{x}\) | \(f'(0)=0\) |
| \(f''(x)=-\cos{x}\) | \(f''(0)=-1\) |
| \(f'''(x)=\sin{x}\) | \(f'''(0)=0\) |
| \(f^{(iv)}(x)=\cos{x}\) | \(f^{(4)}(0)=1\) |
Then, the fourth degree Taylor polynomial around \(x=0\) is
\[\begin{align}T_4(x) &=f(0)+\frac{f'(0)(x-0)}{1!}+\frac{f''(0)(x-0)^{2}}{2!}\\ & \quad+\frac{f'''(0)(x-0)^{3}}{3!}+\frac{f^{(4)}(0)(x-0)^{4}}{4!} \\ &= 1+0-\frac{x^2}{2!}+0+\frac{x^4}{4!} \\ &=1-\frac{x^2}{2!}+\frac{x^4}{4!}.\end{align}\]
So, evaluating at \(x=2\), you have
\[\begin{align} \cos\left(\frac{\pi}{2}\right) &\approx T_4\left(\frac{\pi}{2}\right)\\ &=1-\frac{\frac{\pi}{2}}{2!}+\frac{\frac{\pi}{2}^4}{4!} \\ &=0.02.\end{align}\]
Let's look at another example.
Calculate the value of \(\sqrt{24}\) using a quadratic approximation.
Solution:
In this case, you need to calculate the second degree Taylor polynomial of the function \(g(x)=\sqrt{x}\) since you want a quadratic approximate of \(\sqrt{24}\).
Since Taylor polynomials only allow you to approximate values close to the value at which they are centered, you need a value close to \(24\) where you can actually find the square root easily. So let's take \(25\) since \(\sqrt{25}=5\).
Table 4. Table of derivatives and function values for \(\sqrt{x}\).
| \(g(x)=\sqrt{x}\) | \(g(25)=\sqrt{25}=5\) |
| \(g'(x)=\dfrac{1}{2\sqrt{x}}\) | \(g'(25)=\dfrac{1}{2\sqrt{25}}=\dfrac{1}{10}\) |
| \(g''(x)=-\dfrac{1}{4x^{3/2}}\) | \(g''(25)=-\dfrac{1}{4(25)^{3/2}}=-\dfrac{1}{1000}\) |
Thus, the quadratic (another way of saying second degree) Taylor polynomial of \(\sqrt{x}\) centered at \(x=25\) is
\[T_2(x)=5+\frac{(x-25)}{10}-\frac{(x-25)^2}{1000}.\]
Using the \(T_2(x)\) approximation you get
\[\begin{align} \sqrt{24}&\approx 5+\frac{(24-25)}{10}-\frac{(24-25)^2}{1000} \\ &= 4.899.\end{align}\]
❗❗ Note that the Taylor polynomial calculated in the previous example was not used because it was centered on \(1\), and \(1\) is very far from \(24\), so using it would have given you a very bad approximation.
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How do you find Taylor polynomials?
The formula is based on the derivatives of the function, the center point, and power functions. To see the whole formula, take a look at our Taylor Polynomials article.
What is Taylor's formula used for?
It is used to approximate a complex function with simpler functions and thus estimate the values of the function at points where it is difficult to evaluate.
What is the difference between Taylor series and Taylor polynomial?
A Taylor polynomial takes a fixed number n of derivatives to estimate a function, while a Taylor series takes all the derivatives to estimate a function.
What is the degree of a Taylor polynomial?
The degree of a Taylor polynomial is given by the highest derivative used to estimate the function.
How do you solve Taylor series problems?
Since Taylor series include all the derivatives of a function, it is necessary to find the pattern that the derivatives follow.
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