Let Us Approximate E ^x 1. Approximate E ^0.5 Using Taylor Series. 2. Approximate E ^10 Using Taylor
Understanding the exponential function \( e^x \) is fundamental in calculus, mathematical analysis, and numerous scientific applications. Since the function \( e^x \) can be complex to evaluate directly for arbitrary \( x \), mathematicians often resort to approximation techniques such as Taylor series expansions. These series provide polynomial approximations that can be remarkably accurate within certain intervals. In this article, we will explore how to approximate \( e^x \) at specific points—namely \( x=1 \), \( x=0.5 \), and \( x=10 \)—using Taylor series. We will delve into the derivation of the series, step-by-step calculations, and analyze the accuracy of these approximations.
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Understanding the Taylor Series for \( e^x \)
Definition of Taylor Series
The Taylor series of a function \( f(x) \) centered at a point \( a \) is an infinite sum of terms calculated from the derivatives of \( f \) at \( a \):
\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n
\]
where:
- \( f^{(n)}(a) \) is the \( n \)-th derivative of \( f \) evaluated at \( a \),
- \( n! \) is the factorial of \( n \).
For the exponential function \( e^x \), all derivatives are \( e^x \), which simplifies the process.
Taylor Series for \( e^x \) at \( a=0 \) (Maclaurin Series)
Since \( e^x \) is infinitely differentiable and its derivatives are all \( e^x \), the Taylor series centered at \( a=0 \) (also called the Maclaurin series) is:
\[
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
\]
This series converges for all real \( x \).
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Approximating \( e^x \) at \( x=1 \) and \( x=0.5 \) Using Taylor Series
1. Approximate \( e^1 \) (i.e., \( e \))
The goal is to approximate \( e^1 \) using a finite number of terms from the Taylor series expansion at \( a=0 \):
\[
e^1 \approx \sum_{n=0}^{N} \frac{1^n}{n!}
\]
where \( N \) is the number of terms included.
Calculating the Approximation with Several Terms
Let's compute approximations for \( N=1, 2, 3, 4, 5 \):
- N=1:
\[
e^1 \approx 1 + \frac{1}{1!} = 1 + 1 = 2
\]
- N=2:
\[
e^1 \approx 1 + 1 + \frac{1^2}{2!} = 2 + \frac{1}{2} = 2.5
\]
- N=3:
\[
e^1 \approx 2.5 + \frac{1}{3!} = 2.5 + \frac{1}{6} \approx 2.6667
\]
- N=4:
\[
2.6667 + \frac{1}{4!} = 2.6667 + \frac{1}{24} \approx 2.7083
\]
- N=5:
\[
2.7083 + \frac{1}{5!} = 2.7083 + \frac{1}{120} \approx 2.7167
\]
The actual value of \( e \) is approximately 2.71828, so increasing the number of terms yields a more accurate approximation.
2. Approximate \( e^{0.5} \)
Similarly, for \( x=0.5 \):
\[
e^{0.5} \approx \sum_{n=0}^N \frac{(0.5)^n}{n!}
\]
Calculations for the first few terms:
- N=1:
\[
1 + \frac{0.5}{1!} = 1 + 0.5 = 1.5
\]
- N=2:
\[
1.5 + \frac{(0.5)^2}{2!} = 1.5 + \frac{0.25}{2} = 1.5 + 0.125 = 1.625
\]
- N=3:
\[
1.625 + \frac{(0.5)^3}{3!} = 1.625 + \frac{0.125}{6} \approx 1.625 + 0.0208 = 1.6458
\]
- N=4:
\[
1.6458 + \frac{(0.5)^4}{4!} = 1.6458 + \frac{0.0625}{24} \approx 1.6458 + 0.0026 = 1.6484
\]
- N=5:
\[
1.6484 + \frac{(0.5)^5}{5!} = 1.6484 + \frac{0.03125}{120} \approx 1.6484 + 0.00026 = 1.6487
\]
The true value of \( e^{0.5} \) is approximately 1.64872, indicating that with five terms, the approximation is quite close.
Analysis of Approximation Accuracy
The Taylor series converges rapidly for these values, especially near the center point \( a=0 \). The accuracy improves as more terms are included. However, the number of terms needed depends on the magnitude of \( x \); larger \( |x| \) generally requires more terms for the same accuracy.
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Approximating \( e^{10} \) Using Taylor Series
Challenges with Direct Taylor Series at \( a=0 \)
Attempting to approximate \( e^{10} \) directly using the Maclaurin series:
\[
e^{10} \approx \sum_{n=0}^N \frac{10^n}{n!}
\]
requires a very large \( N \) because the terms grow significantly before factorial growth dominates, and the series converges slowly for large \( x \).
Practical Approach: Use of Series Expansion at a Closer Center
To improve convergence, we can choose a different center \( a \) close to 10, such as \( a=10 \), and expand \( e^{x} \) around \( x=10 \):
\[
e^{x} = e^{10} + e^{10} (x - 10) + \frac{e^{10}}{2!} (x - 10)^2 + \dots
\]
But since \( e^{x} \) is exponential, the Taylor expansion at \( a=10 \) simplifies to:
\[
e^{x} = e^{10} \sum_{n=0}^{\infty} \frac{(x - 10)^n}{n!}
\]
If we want to approximate \( e^{10} \), setting \( x=10 \):
\[
e^{10} \approx e^{10} \times 1 = e^{10}
\]
which is trivial. To approximate \( e^{10} \) via Taylor series, a better approach is to use the property:
\[
e^{10} = (e^{1})^{10}
\]
and approximate \( e^{1} \) or \( e^{0.1} \) and then raise to the 10th power, or use properties like:
\[
e^{x} = \left( e^{x/n} \right)^n
\]
for some \( n \).
Approximate \( e^{10} \) via Repeated Use of \( e^{1} \)
Suppose we choose \( n=10 \):
\[
e^{10} = (e^{1})^{10}
\]
Using the approximation for \( e^{1} \) from earlier (say, with 5 terms: approximately 2.7183), we compute:
\[
e^{10} \approx (2.7183)^{10}
\]
This can be calculated using logarithms or iterative multiplication.
Alternatively, directly approximating \( e^{10} \) with Taylor series at \( a=0 \):
- Number of terms needed: Since \( 10^n / n! \) grows initially, then decreases, the dominant term occurs around \( n=10 \), because:
\[
\frac{10^{10}}{10!} \approx \frac{10^{10}}{3
