Problem 3
Question
Consider any function \(f(x)\) defined for \(a \leqslant x \leqslant b\). Approximate this function by a constant. Show that the best such constant (in the mean-square sense, i.e., minimizing the mean-square deviation) is the constant equal to the average of \(f(x)\) over the interval \(a \leqslant x \leqslant b\).
Step-by-Step Solution
Verified Answer
The value of the constant \(C\) that minimizes the mean square deviation of \(f(x)\) over the range \([a, b]\) is exactly equal to the average value of \(f(x)\) over this range, given by \(C = \frac{1}{b-a} \int_{a}^{b} f(x) dx.\)
1Step 1: Set up the integral for the mean squared error.
Over the interval \([a, b]\), the mean square deviation of the constant \(C\) from the function \(f(x)\) is given by \[\int_{a}^{b} (f(x) - C)^2 dx.\]The task involves finding the value of \(C\) that minimizes this quantity.
2Step 2: Calculate the derivative and find its minimum.
Next step, differentiate the integral with respect to \(C\) and set the result equal to zero.The derivative of \(\int_{a}^{b} (f(x)- C)^2 dx\) with respect to \(C\) is \(-2\int_{a}^{b} (f(x)- C) dx\).Setting this equal to 0, yields\[-2\int_{a}^{b} (f(x) - C) dx = 0,\]or\[\int_{a}^{b} f(x) dx = \int_{a}^{b} C dx\]It implies that \(C(b-a) = \int_{a}^{b} f(x) dx\).Therefore, the constant \(C\) that minimize the mean square deviation is then the average value of \(f(x)\) over \([a, b]\), i.e., \[C = \frac{1}{b-a} \int_{a}^{b} f(x) dx.\]
3Step 3: Proof
The second derivative of the integral is \(2\int_{a}^{b} dx\), which is always greater than 0, so \(C\) is indeed a minimum value.
Key Concepts
Mean-Square DeviationAverage Value of FunctionIntegral Calculus
Mean-Square Deviation
When approximating a function by a constant over a given interval, one effective method is to minimize the mean-square deviation. The mean-square deviation measures how much the function values, on average, deviate from the constant we choose. To find this, we examine the integral \[ \int_{a}^{b} (f(x) - C)^2 dx, \]where \(C\) is the constant approximation of the function \(f(x)\) over the interval \([a, b]\). This integral represents the total squared error – essentially the sum of squared differences between \(f(x)\) and \(C\) over the interval.
By differentiating this integral with respect to \(C\), we determine how the error changes as we modify \(C\). The aim is to find where this error is smallest, which occurs when the derivative equals zero, indicating the error is at a minimum point. We use this technique to ensure that our approximation is as close as possible to the actual function in the "mean-square" sense.
By differentiating this integral with respect to \(C\), we determine how the error changes as we modify \(C\). The aim is to find where this error is smallest, which occurs when the derivative equals zero, indicating the error is at a minimum point. We use this technique to ensure that our approximation is as close as possible to the actual function in the "mean-square" sense.
Average Value of Function
To find the best constant approximation for a function \(f(x)\) using mean-square deviation, it turns out that the optimal constant \(C\) is the average value of the function over the interval \([a, b]\). But how do we calculate this average value?
The average value of a function over an interval \([a, b]\) is given by the formula \[C = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx, \]which represents the integral of the function divided by the length of the interval.
The average value of a function over an interval \([a, b]\) is given by the formula \[C = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx, \]which represents the integral of the function divided by the length of the interval.
- This formula effectively sums up the function values across the interval and normalizes by the interval's length, providing the "mean" or average value.
- When you use this average value as the constant \(C\), you minimize the mean-square deviation, making it the best possible approximation with regards to this measure.
Integral Calculus
Integral calculus plays a crucial role in finding both mean-square deviations and average values of functions. The entire process of approximating a function by a constant using mean-square deviation relies on the integration of functions over specific intervals.
Firstly, let's understand integration in this context. Integration helps determine the area under the curve for a given function \(f(x)\) on an interval \([a, b]\). It aggregates values of the function, crucial for evaluating deviations or finding averages.
Firstly, let's understand integration in this context. Integration helps determine the area under the curve for a given function \(f(x)\) on an interval \([a, b]\). It aggregates values of the function, crucial for evaluating deviations or finding averages.
- We use integrals like \[\int_{a}^{b} (f(x) - C)^2 dx, \]to calculate the mean-square deviation. This integral gives us a measure of error for approximation.
- Moreover, \[\int_{a}^{b} f(x) \, dx \]allows us to determine the total values of the function across the interval, facilitating the calculation of average values.
Other exercises in this chapter
Problem 2
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Prove that \((5.6 .10)\) is valid in the following way. Assume \(L(u) / \sigma\) is piecewise smooth so that $$ \frac{L(u)}{\sigma}=\sum_{n=1}^{\infty} b_{n} \p
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