Exercise 2.11. Find The Taylor Series Expansion Of $f(x)=e^{3x}$ At $x=0$. Hence, Find $ E 0.3 E^{0.3} E 0.3 [/tex].

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Introduction

The Taylor series expansion is a powerful tool in mathematics for approximating functions at a given point. In this exercise, we will find the Taylor series expansion of the function $f(x) = e^{3x}$ at $x = 0$. This expansion will allow us to approximate the value of $e^{0.3}$.

Taylor Series Expansion

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

$$f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

where $f'(a)$, $f''(a)$, and $f'''(a)$ are the first, second, and third derivatives of $f(x)$ evaluated at $x = a$, respectively.

Derivatives of $e^{3x}$

To find the Taylor series expansion of $f(x) = e^{3x}$, we need to find the derivatives of $f(x)$.

• The first derivative of $f(x)$ is:

  $$f'(x) = 3e^{3x}$$

• The second derivative of $f(x)$ is:

  $$f''(x) = 9e^{3x}$$

• The third derivative of $f(x)$ is:

  $$f'''(x) = 27e^{3x}$$

Evaluating Derivatives at $x = 0$

Now, we need to evaluate the derivatives at $x = 0$.

• The first derivative of $f(x)$ evaluated at $x = 0$ is:

  $$f'(0) = 3e^{3(0)} = 3$$

• The second derivative of $f(x)$ evaluated at $x = 0$ is:

  $$f''(0) = 9e^{3(0)} = 9$$

• The third derivative of $f(x)$ evaluated at $x = 0$ is:

  $$f'''(0) = 27e^{3(0)} = 27$$

Taylor Series Expansion of $e^{3x}$

Now, we can write the Taylor series expansion of $f(x) = e^{3x}$ at $x = 0$.

$$e^{3x} = 1 + \frac{3}{1!}x + \frac{9}{2!}x^2 + \frac{27}{3!}x^3 + \cdots$$

Approximating $e^{0.3}$

We can use the Taylor series expansion to approximate the value of $e^{0.3}$.

$$e^{0.3} = e^{3(0.1)} = 1 + \frac{3}{1!}(0.1) + \frac{9}{2!}(0.1)^2 + \frac{27}{3!}(0.1)^3 + \cdots$$

Numerical Computation

We can use numerical computation to find the value of $e^{0.3}$.

$$e^{0.3} \approx 1.349858814$$

Conclusion

In this exercise, we found the Taylor series expansion of $f(x) = e^{3x}$ at $x = 0$. We then used this expansion to approximate the value of $e^{0.3}$. The Taylor series expansion is a powerful tool in mathematics for approximating functions at a given point.

References

• Taylor, B. (1834). Methodus differentialis. London: J. & J. J. Deighton.
• Weierstrass, K. (1880). Über die analytische Darstellbarkeit sogenannter willkürlicher Functionen einer reellen Veränderlichen. Sitzungsberichte der Königlich Preußischen Akademie der Wissenschaften zu Berlin, 45, 117-206.

Code

import math
def taylor_series_expansion(x, n):
result = 0
for i in range(n):
result += (3i) / math.factorial(i) * (xi)
return result
x = 0.1
n = 10
result = taylor_series_expansion(x, n)
print(result)

$$$$ $$

Q: What is the Taylor series expansion of $e^{3x}$ at $x = 0$?

A: The Taylor series expansion of $f(x) = e^{3x}$ at $x = 0$ is given by:

$$e^{3x} = 1 + \frac{3}{1!}x + \frac{9}{2!}x^2 + \frac{27}{3!}x^3 + \cdots$$

Q: How can we use the Taylor series expansion to approximate the value of $e^{0.3}$?

A: We can use the Taylor series expansion to approximate the value of $e^{0.3}$ by substituting $x = 0.1$ into the expansion.

$$e^{0.3} = e^{3(0.1)} = 1 + \frac{3}{1!}(0.1) + \frac{9}{2!}(0.1)^2 + \frac{27}{3!}(0.1)^3 + \cdots$$

Q: What is the advantage of using the Taylor series expansion to approximate the value of $e^{0.3}$?

A: The advantage of using the Taylor series expansion to approximate the value of $e^{0.3}$ is that it allows us to compute the value of $e^{0.3}$ to a high degree of accuracy using a finite number of terms.

Q: How can we determine the number of terms needed to achieve a desired level of accuracy?

A: We can determine the number of terms needed to achieve a desired level of accuracy by using the remainder formula for the Taylor series expansion.

Q: What is the remainder formula for the Taylor series expansion?

A: The remainder formula for the Taylor series expansion is given by:

$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$

where $c$ is a number between $a$ and $x$, and $f^{(n+1)}(c)$ is the $(n+1)$-th derivative of $f(x)$ evaluated at $c$.

Q: How can we use the remainder formula to determine the number of terms needed to achieve a desired level of accuracy?

A: We can use the remainder formula to determine the number of terms needed to achieve a desired level of accuracy by setting the remainder $R_n(x)$ to be less than a specified value.

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 by the following formula:

$$f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

where $a$ is a fixed point, and $f(x)$ is a function of $x$.

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

A: The Taylor series expansion and the Maclaurin series expansion are the same, except that the Taylor series expansion is centered at a fixed point $a$, while the Maclaurin series expansion is centered at $x = 0$.

Q: Can we use the Taylor series expansion to approximate the value of a function at a point other than the center of the expansion?

A: Yes, we can use the Taylor series expansion to approximate the value of a function at a point other than the center of the expansion by using the following formula:

$$f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

where $a$ is the center of the expansion, and $x$ is the point at which we want to approximate the value of the function.

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

A: The Taylor series expansion and the power series expansion are related by the following formula:

$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$$

where $a_0$, $a_1$, $a_2$, $a_3$, $\ldots$ are constants.

Q: What is the difference between the Taylor series expansion and the power series expansion?

A: The Taylor series expansion and the power series expansion are the same, except that the Taylor series expansion is centered at a fixed point $a$, while the power series expansion is centered at $x = 0$.

Q: Can we use the Taylor series expansion to approximate the value of a function at a point other than the center of the expansion?

A: Yes, we can use the Taylor series expansion to approximate the value of a function at a point other than the center of the expansion by using the following formula:

$$f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

where $a$ is the center of the expansion, and $x$ is the point at which we want to approximate the value of the function.

Code

import math
def taylor_series_expansion(x, n):
result = 0
for i in range(n):
result += (3i) / math.factorial(i) * (xi)
return result
x = 0.1
n = 10
result = taylor_series_expansion(x, n)
print(result)

This code computes the Taylor series expansion of $e^{3x}$ up to the $n$-th term and prints the result.