Problem 2 - A Primer of Real Analysis

Question

Use the 5 th order Taylor polynomial for \(f(x)=\sqrt{x}\) at 1 to estimate \(\sqrt{1.2}\). Is this an underestimate or an overestimate? Find an upper bound for the largest amount by which the estimate and \(\sqrt{1.2}\) differ.

Step-by-Step Solution

Answer

Estimate is about 1.095445104; it's an overestimate with error ≤ 0.000065.

Step 1: Understand the Taylor Series

The Taylor series for a function \( f(x) \) centered at \( a \) is given by:\[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]We need to find the 5th order Taylor polynomial for \( f(x) = \sqrt{x} \) centered at \( a = 1 \).

Step 2: Calculate Derivatives of f(x) = x^(1/2)

Calculate the first few derivatives of \( f(x) = x^{1/2} \):- \( f(x) = x^{1/2} \)- \( f'(x) = \frac{1}{2}x^{-1/2} \)- \( f''(x) = -\frac{1}{4}x^{-3/2} \)- \( f'''(x) = \frac{3}{8}x^{-5/2} \)- \( f^{(4)}(x) = -\frac{15}{16}x^{-7/2} \)- \( f^{(5)}(x) = \frac{105}{32}x^{-9/2} \)

Step 3: Evaluate Derivatives at a = 1

Evaluate each of these derivatives at \( x = 1 \):- \( f(1) = 1 \)- \( f'(1) = \frac{1}{2} \)- \( f''(1) = -\frac{1}{4} \)- \( f'''(1) = \frac{3}{8} \)- \( f^{(4)}(1) = -\frac{15}{16} \)- \( f^{(5)}(1) = \frac{105}{32} \)

Step 4: Construct the 5th Order Taylor Polynomial

Substitute these values into the Taylor series formula:\[T_5(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} - \frac{15}{16}\frac{(x-1)^4}{4!} + \frac{105}{32}\frac{(x-1)^5}{5!}\]

Step 5: Estimate \(\sqrt{1.2}\) Using the Taylor Polynomial

Substitute \( x = 1.2 \) into the polynomial:\[T_5(1.2) = 1 + \frac{1}{2}(0.2) - \frac{1}{4}\frac{(0.2)^2}{2} + \frac{3}{8}\frac{(0.2)^3}{6} - \frac{15}{16}\frac{(0.2)^4}{24} + \frac{105}{32}\frac{(0.2)^5}{120}\]Calculate this to obtain:\( T_5(1.2) \approx 1.095445104 \)

Step 6: Determine Underestimate or Overestimate

The function \( f(x) = \sqrt{x} \) is concave down for \( x > 0 \) (since \( f''(x) < 0 \)). Therefore, the Taylor polynomial will be an overestimate for \( \sqrt{1.2} \).

Step 7: Find the Upper Bound of the Error

The error in the Taylor estimate \( R_n(x) \) is bounded by the next term in the series:\[|R_5(1.2)| \leq \frac{|f^{(6)}(c)|}{6!}|x-a|^6\]Where \(c\) is some number between \(1\) and \(1.2\). We approximate \( f^{(6)}(x) = -\frac{945}{64}x^{-11/2} \) and estimate over the interval \([1, 1.2]\) to find the bound. The maximum occurs at \(x = 1\), thus:\[|R_5(1.2)| \leq \frac{945}{64 \times 720} \times (0.2)^6 \approx 0.000065\]

Step 8: Final Step: Conclusion

The estimate for \( \sqrt{1.2} \) is approximately \( 1.095445104 \), which is an overestimate. The maximum possible error in this estimate is approximately \( 0.000065 \).

Key Concepts

Taylor polynomial

When we talk about a Taylor polynomial, we are really diving into one of the foundational concepts in calculus. A Taylor polynomial is a tool that helps us approximate a function at a certain point using derivatives. Think of it as a bridge between a complex function and a much simpler polynomial that behaves similarly near a specific point.

For any given function, you can construct a Taylor polynomial by following these general steps:

• Find the derivatives of the function at the point where you want the approximation (these are called "center points").
• Use these derivatives to build the polynomial that best approximates the function in the neighborhood of the center point.
• The polynomial will include terms up to the degree that you choose (like the 5th degree in our exercise).

The formula for the nth-degree Taylor polynomial of a function, centered at some point \(a\), looks like this:\[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \cdots + \frac{f^{(n)}(a)(x-a)^n}{n!}\]This polynomial gives a good approximation to the function as you calculate further terms. However, keep in mind, the more terms you include, the better the approximation, especially as you move away from the center point.

error estimation

Estimating the error in a Taylor polynomial approximation is crucial to understand how off our approximation might be. The error, often denoted by \(R_n(x)\), is the difference between the true value of the function and the estimated value given by the polynomial.

To get a grasp on the error size, we use something called the "Lagrange error bound," which provides an upper bound for the error term. This bound helps us understand the "worst-case" scenario of how much our approximation might deviate from the actual function value.

Mathematically, the error bound is expressed as:\[|R_n(x)| \leq \frac{|f^{(n+1)}(c)|}{(n+1)!}|x-a|^{n+1}\]Here, \(c\) is some number between our center point \(a\) and the value at which we are estimating, \(x\). In practical terms, this error estimation allows us to say, "Although we approximate the value, it won't stray too far from the actual function." This can be extremely helpful when accuracy is of significant concern.

concave down

The concept of "concave down" is essential when evaluating whether a Taylor polynomial serves as an underestimate or overestimate. A function is said to be concave down when it curves downward, resembling an upside-down cup. This is visually represented by the function opening downwards.

To confirm if a function is concave down, you look at the function's second derivative, \(f''(x)\). If the second derivative is negative over an interval, the function is concave down on that interval.

• When a function is concave down, any tangent lines (and similarly its Taylor polynomial estimates) will lie above the curve.
• This typically indicates that your polynomial is giving an overestimate of the function's true value at a given point.

For instance, in our case with \(f(x) = \sqrt{x}\), since the second derivative \(f''(x)\) is negative for the values of \(x > 0\), the function is indeed concave down. Thus, our Taylor polynomial overestimates \(\sqrt{1.2}\). Understanding this relationship helps refine our expectations of polynomial approximations and guide us toward recognizing when our estimates are consistently above or below the actual function.