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When \(0 \leq h \leq 0.01,\) show that \(e^{h}\) may be replaced by \(1+h\) with an error of magnitude no greater than 0.6\(\%\) of \(h .\) Use \(e^{0.01}=1.01 .\)

Short Answer

Expert verified
The error is less than 0.6% of \( h \) for \( 0 \leq h \leq 0.01 \).

Step by step solution

01

Understand the Exponential Approximation

For small values of a variable, like in this case where \( 0 \leq h \leq 0.01 \), an exponential function can be approximated using a series expansion. The approximation \( e^h \approx 1 + h \) is derived from the Taylor series: \[ e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \ldots \]For sufficiently small \( h \), higher-order terms \( (h^2, h^3, \ldots) \) become negligible.
02

Evaluate at Upper Bound

At the upper bound \( h = 0.01 \), calculate the actual value \( e^h = e^{0.01}. \) This is approximated already as \( e^{0.01} = 1.01 \). The linear approximation gives \( 1 + 0.01 = 1.01 \). Hence, the linear approximation perfectly matches at this point.
03

Calculate the Error Term

For the approximation \( e^h \approx 1 + h \), the error term can be calculated using: \[ \text{Error} = \left| e^h - (1+h) \right| = \left| \frac{h^2}{2!} + \frac{h^3}{3!} + \ldots \right| \]We are asked if this error is no greater than 0.6% of \( h \). We calculate the error for \( h = 0.01 \) using the first non-constant term: \[ \text{Error} \approx \frac{(0.01)^2}{2} = \frac{0.0001}{2} = 0.00005 \]
04

Check Against Error Percentage Constraint

We need the error to be \( \leq 0.6\% \) of \( h \), which is \[ 0.006 \times 0.01 = 0.00006 \] Since the calculated error at \( h = 0.01 \) is \( 0.00005 \), which is less than \( 0.00006 \), the constraint is satisfied, proving that the error is no more than 0.6\% of \( h \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Approximation
When we talk about exponential approximation, we're referring to how we can use simpler expressions to represent more complex exponential functions. This is particularly useful when dealing with small numbers, as it allows calculations to be quicker and easier. In this context, when the variable \( h \) is very small, we use the linear approximation of the exponential function:
  • The general idea is derived from the Taylor series expansion for \( e^h \): \[ e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \ldots\]
  • For very small \( h \), such as when \( 0 \leq h \leq 0.01 \), the higher-order terms \( (h^2, h^3, \ldots) \) are much smaller compared to the initial terms and can be ignored. Thus, the approximation \( e^h \approx 1 + h \) becomes valid. This simplification comes in handy when performing quick estimations.
Think of it as zooming in on the curve of \( e^h \) at very small \( h \). At these points, the curve is almost a straight line, and the line \( 1 + h \) aligns very closely with \( e^h \). This knowledge is especially useful in fields requiring rapid computations without compromising accuracy significantly.
Error Analysis
Error analysis helps us understand how far off our approximations might be from the true values. In the given exercise, we are checking whether replacing \( e^h \) with \( 1 + h \) yields a negligible error within a specific limit.Let's break this down:
  • We calculate the error term using the formula: \[ \text{Error} = \left| e^h - (1 + h) \right| = \left| \frac{h^2}{2!} + \frac{h^3}{3!} + \ldots \right|\] This formula accounts for the terms that we initially discarded during approximation.
  • For \( h = 0.01 \), the significant error is primarily due to the first neglected term, \( \frac{h^2}{2} \).
The critical part of this analysis is to ensure that this error does not exceed a reasonable threshold. In this particular task, we need the error to be not more than 0.6% of \( h \). This allows us to ensure that the approximation is about as close as we need it to be for effective applications, retaining a manageable level of accuracy.
Mathematical Proof
Mathematical proof is the heart of determining whether our approximations and assumptions hold true. In exercises like these, we delve into the proof by establishing conditions and verifying them against given constraints:
  • We start by proving that the linear approximation \( 1 + h \) is valid by calculating both the approximation and the actual value at the upper limit \( h = 0.01 \). We found both to be exactly \( 1.01 \), demonstrating the approximation's accuracy at this point.
  • Next, we validate the error bounds. We need to ensure that the approximation error is within 0.6% of \( h \). We compute: \[ \text{Maximum error allowed} = 0.006 \times 0.01 = 0.00006\]Then verify: \[ \text{Actual error} = \frac{(0.01)^2}{2} = 0.00005\]
  • Since the actual error is \( 0.00005 \), which is less than \( 0.00006 \), the approximation holds under the set condition. This step-by-step check provides the proof needed to assert the reliability of \( 1 + h \) as an approximation of \( e^h \) within the specified range.
The proof underscores the importance of checking each assumption and calculation, confirming that mathematical shortcuts do not compromise accuracy beyond acceptable limits.

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