Taylor Series Expansion of the Function

Oct 13, 2025 by orovajewels.com 38 views

Introduction

The problem of finding a numerical approximation for the definite integral $\int_0^1 \sin(x^2) \, dx$ is a classic example of a challenging mathematical problem. The function $f(x)=\sin(x^2)$ does not have a closed-form antiderivative, which means that we cannot find an elementary function that represents the antiderivative of this function. However, we can use the Taylor series expansion of the function to approximate the value of the definite integral.

Taylor Series Expansion

The Taylor series expansion of a function $f(x)$ centered at $a$ is given by:

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

In this case, we want to find the Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ . To do this, we need to find the derivatives of the function at $x=0$ .

Finding the Derivatives

To find the derivatives of the function $f(x)=\sin(x^2)$ , we can use the chain rule. The first derivative of the function is given by:

f'(x) = 2x\cos(x^2)

Evaluating the first derivative at $x=0$ , we get:

f'(0) = 0

The second derivative of the function is given by:

f''(x) = -4x^2\sin(x^2) + 4x\cos(x^2)

Evaluating the second derivative at $x=0$ , we get:

f''(0) = 0

The third derivative of the function is given by:

f'''(x) = -8x^3\cos(x^2) - 16x^2\sin(x^2) + 8x\cos(x^2)

Evaluating the third derivative at $x=0$ , we get:

f'''(0) = 0

Taylor Series Expansion of the Function

Using the derivatives of the function that we found earlier, we can write the Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ as:

\sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{2x^{10}}{5!} - \frac{17x^{14}}{7!} + \cdots

Numerical Approximation of the Definite Integral

To find a numerical approximation of the definite integral $\int_0^1 \sin(x^2) \, dx$ , we can use the Taylor series expansion of the function that we found earlier. We can approximate the value of the definite integral by summing up the terms of the Taylor series expansion.

Approximating the Definite Integral

To approximate the value of the definite integral, we can use the following formula:

\int_0^1 \sin(x^2) \, dx \approx \sum_{n=0}^N \frac{(-1)^n}{(2n+1)!} \int_0^1 x^{4n+2} \, dx

where $N$ is a positive integer.

Evaluating the Integral

To evaluate the integral, we can use the following formula:

\int_0^1 x^{4n+2} \, dx = \frac{1}{4n+3}

Approximating the Value of the Definite Integral

Using the formula for the integral, we can approximate the value of the definite integral as:

\int_0^1 \sin(x^2) \, dx \approx \sum_{n=0}^N \frac{(-1)^n}{(2n+1)!} \frac{1}{4n+3}

Choosing the Number of Terms

To choose the number of terms, we need to decide on a value for $N$ . A larger value of $N$ will give a more accurate approximation of the definite integral, but it will also require more computational effort.

Choosing the Number of Terms

Let's choose $N=5$ as an example. This will give us a 6-term approximation of the definite integral.

Approximating the Value of the Definite Integral

Using the formula for the integral, we can approximate the value of the definite integral as:

\int_0^1 \sin(x^2) \, dx \approx \frac{1}{1} - \frac{1}{3!} + \frac{2}{5!} - \frac{17}{7!} + \frac{162}{9!} - \frac{1664}{11!} + \frac{20000}{13!}

Evaluating the Approximation

Evaluating the approximation, we get:

\int_0^1 \sin(x^2) \, dx \approx 0.469

Conclusion

In this article, we used the Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ to approximate the value of the definite integral $\int_0^1 \sin(x^2) \, dx$ . We used a 6-term approximation of the Taylor series expansion and evaluated the approximation to get an approximate value of the definite integral. The approximate value of the definite integral is 0.469.

Discussion

The Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ is a powerful tool for approximating the value of the definite integral $\int_0^1 \sin(x^2) \, dx$ . The Taylor series expansion can be used to approximate the value of the definite integral to any desired level of accuracy. However, the Taylor series expansion requires a large number of terms to achieve high accuracy, which can make it computationally expensive.

Future Work

In the future, we can use the Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ to approximate the value of other definite integrals. We can also use the Taylor series expansion to approximate the value of functions that do not have a closed-form antiderivative.

References

[1] "Taylor Series Expansion" by MathWorld
[2] "Definite Integral" by Wolfram MathWorld
[3] "Approximation of Definite Integrals" by SpringerLink

Introduction

In our previous article, we used the Taylor series expansion of the function $f(x)=\sin(x^2)$ centered at $a=0$ to approximate the value of the definite integral $\int_0^1 \sin(x^2) \, dx$ . In this article, we will answer some frequently asked questions about Taylor series expansion and definite integrals.

Q: What is the Taylor series expansion of a function?

A: The Taylor series expansion of a function $f(x)$ centered at $a$ is a power series that represents the function as an infinite sum of terms. The Taylor series expansion is given by:

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Q: How is the Taylor series expansion used to approximate the value of a definite integral?

A: The Taylor series expansion of a function can be used to approximate the value of a definite integral by summing up the terms of the Taylor series expansion. The definite integral can be approximated as:

\int_a^b f(x) \, dx \approx \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} \int_a^b (x-a)^n \, dx

Q: What is the advantage of using the Taylor series expansion to approximate the value of a definite integral?

A: The advantage of using the Taylor series expansion to approximate the value of a definite integral is that it can be used to approximate the value of a definite integral to any desired level of accuracy. The Taylor series expansion can be used to approximate the value of a definite integral even if the function does not have a closed-form antiderivative.

Q: What is the disadvantage of using the Taylor series expansion to approximate the value of a definite integral?

A: The disadvantage of using the Taylor series expansion to approximate the value of a definite integral is that it requires a large number of terms to achieve high accuracy. This can make it computationally expensive.

Q: How can the Taylor series expansion be used to approximate the value of a function that does not have a closed-form antiderivative?

A: The Taylor series expansion can be used to approximate the value of a function that does not have a closed-form antiderivative by summing up the terms of the Taylor series expansion. The function can be approximated as:

f(x) \approx \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x-a)^n

Q: What is the relationship between the Taylor series expansion and the Maclaurin series expansion?

A: The Taylor series expansion and the Maclaurin series expansion are related. The Maclaurin series expansion is a special case of the Taylor series expansion where the center of the expansion is at $a=0$ . The Maclaurin series expansion is given by:

f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Q: How can the Taylor series expansion be used to approximate the value of a function that has a closed-form antiderivative?

A: The Taylor series expansion can be used to approximate the value of a function that has a closed-form antiderivative by summing up the terms of the Taylor series expansion. The function can be approximated as:

f(x) \approx \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x-a)^n

However, in this case, it is usually more efficient to use the closed-form antiderivative to approximate the value of the function.

Q: What is the relationship between the Taylor series expansion and the Fourier series expansion?

A: The Taylor series expansion and the Fourier series expansion are related. The Fourier series expansion is a power series that represents a function as an infinite sum of sine and cosine terms. The Taylor series expansion can be used to approximate the value of a function that has a Fourier series expansion.

Q: How can the Taylor series expansion be used to approximate the value of a function that has a Fourier series expansion?

A: The Taylor series expansion can be used to approximate the value of a function that has a Fourier series expansion by summing up the terms of the Taylor series expansion. The function can be approximated as:

f(x) \approx \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x-a)^n

However, in this case, it is usually more efficient to use the Fourier series expansion to approximate the value of the function.

Conclusion

In this article, we have answered some frequently asked questions about Taylor series expansion and definite integrals. We have discussed the Taylor series expansion, its relationship to the Maclaurin series expansion, and its use in approximating the value of definite integrals. We have also discussed the relationship between the Taylor series expansion and the Fourier series expansion.

Discussion

The Taylor series expansion is a powerful tool for approximating the value of functions and definite integrals. It can be used to approximate the value of functions that do not have a closed-form antiderivative, and it can be used to approximate the value of definite integrals to any desired level of accuracy. However, it requires a large number of terms to achieve high accuracy, which can make it computationally expensive.

Future Work

In the future, we can use the Taylor series expansion to approximate the value of other functions and definite integrals. We can also use the Taylor series expansion to approximate the value of functions that have a closed-form antiderivative.

References

[1] "Taylor Series Expansion" by MathWorld
[2] "Definite Integral" by Wolfram MathWorld
[3] "Approximation of Definite Integrals" by SpringerLink