Problem 6
Question
In this exercise we investigation the Taylor series of polynomial functions. a. Find the 3 rd order Taylor polynomial centered at \(a=0\) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Does your answer surprise you? Explain. b. Without doing any additional computation, find the 4 th, 12 th, and 100 th order Taylor polynomials (centered at \(a=0\) ) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Why should you expect this? c. Now suppose \(f(x)\) is a degree \(m\) polynomial. Completely describe the \(n\) th order Taylor polynomial (centered at \(a=0\) ) for each \(n\).
Step-by-Step Solution
Verified Answer
The 3rd order Taylor polynomial is \( -1 + 3x - 2x^2 + x^3 \). Higher order Taylor polynomials are the same as the original polynomial. For any degree \( m \) polynomial, the \( n \)-th order Taylor polynomial \( n \ge m \) equals the original polynomial.
1Step 1: Identify the Function and its Derivatives
Given the function is \(f(x) = x^3 - 2x^2 + 3x - 1\). Calculate the derivatives needed for the 3rd order Taylor polynomial centered at \(a=0\). \(f(0) = -1\)\(f'(x) = 3x^2 - 4x + 3\)\(f'(0) = 3\)\(f''(x) = 6x - 4\)\(f''(0) = -4\)\(f'''(x) = 6\)\(f'''(0) = 6\)
2Step 2: Write the Taylor Polynomial Formula
The Taylor polynomial of order \( n \) centered at \( a \) is given by:\[ T_n(x) = f(a) + f'(a)\frac{(x-a)}{1!} + f''(a)\frac{(x-a)^2}{2!} + f'''(a)\frac{(x-a)^3}{3!} + \,\dots+ f^{(n)}(a)\frac{(x-a)^n}{n!} \]
3Step 3: Substitute the Values
Substitute \(a = 0\) and the calculated values into the Taylor polynomial formula:\[ T_3(x) = f(0) + f'(0)\frac{x}{1!} + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} \]\[ T_3(x) = -1 + 3x - 2x^2 + x^3 \]
4Step 4: Verify the Result
Observe that the 3rd order Taylor polynomial is the same as the original function, i.e., \( T_3(x) = f(x) \).
5Step 5: Higher Order Taylor Polynomials
Since the function \( f(x) = x^3 - 2x^2 + 3x - 1\) is a polynomial of degree 3, all derivatives of order higher than 3 are zero. Thus, the 4th, 12th, and 100th order Taylor polynomials will all be identical to the 3rd order polynomial. This means: \( T_4(x) = T_{12}(x) = T_{100}(x) = x^3 - 2x^2 + 3x - 1 \)
6Step 6: General Case for Any Order Polynomial
For any degree \( m \) polynomial \( f(x) \), the \( n \)-th order Taylor polynomial centered at \( a=0 \) is simply the original polynomial \( f(x) \) for any \( n \ge m \). This is because all derivatives of order higher than \( m \) are zero.
Key Concepts
Taylor polynomialDerivativesPolynomial function analysisCalculus
Taylor polynomial
A Taylor polynomial is a way to approximate more complex functions using polynomials. For a given function, you can build a Taylor polynomial of order n centered at _a_, which is expressed as:
\ T_n(x) = f(a) + f'(a)\frac{(x-a)}{1!} + f''(a)\frac{(x-a)^2}{2!} + f'''(a)\frac{(x-a)^3}{3!} + \dots+ f^{(n)}(a)\frac{(x-a)^n}{n!} \
This means you need to calculate the derivatives of the function and evaluate them at the point _a_. By substituting these values back into the polynomial, you get a Taylor polynomial that approximates your original function around the point _a_.
In the exercise above, we found the 3rd order Taylor polynomial for the function \(f(x) = x^3 - 2x^2 + 3x - 1\), centered at \(a = 0\). This gives us a polynomial \(T_3(x)\) that matches the original function. Since the original function is already a polynomial of degree 3, its 3rd, 4th, 12th, and 100th order Taylor polynomials are all the same and equal to \(f(x)\).
\ T_n(x) = f(a) + f'(a)\frac{(x-a)}{1!} + f''(a)\frac{(x-a)^2}{2!} + f'''(a)\frac{(x-a)^3}{3!} + \dots+ f^{(n)}(a)\frac{(x-a)^n}{n!} \
This means you need to calculate the derivatives of the function and evaluate them at the point _a_. By substituting these values back into the polynomial, you get a Taylor polynomial that approximates your original function around the point _a_.
In the exercise above, we found the 3rd order Taylor polynomial for the function \(f(x) = x^3 - 2x^2 + 3x - 1\), centered at \(a = 0\). This gives us a polynomial \(T_3(x)\) that matches the original function. Since the original function is already a polynomial of degree 3, its 3rd, 4th, 12th, and 100th order Taylor polynomials are all the same and equal to \(f(x)\).
Derivatives
To construct a Taylor polynomial effectively, you need to know how to compute the derivatives of the function. It's essential to understand both the concept of derivatives and how to find them.
In simple terms, the derivative of a function represents the rate at which the function's value changes as its input changes.
For a function \(f(x) = x^3 - 2x^2 + 3x - 1\), the derivatives are:
Knowing these principles is crucial in Taylor polynomial constructions since each term of the Taylor polynomial involves a derivative of the function. The higher the order of the Taylor polynomial, the more derivatives you include.
In simple terms, the derivative of a function represents the rate at which the function's value changes as its input changes.
For a function \(f(x) = x^3 - 2x^2 + 3x - 1\), the derivatives are:
- First derivative: \(f'(x) = 3x^2 - 4x + 3\), evaluated as \(f'(0) = 3\)
- Second derivative: \(f''(x) = 6x - 4\), evaluated as \(f''(0) = -4\)
- Third derivative: \(f'''(x) = 6\), evaluated as \(f'''(0) = 6\)
Knowing these principles is crucial in Taylor polynomial constructions since each term of the Taylor polynomial involves a derivative of the function. The higher the order of the Taylor polynomial, the more derivatives you include.
Polynomial function analysis
Analyzing polynomial functions helps in understanding the behavior of Taylor polynomials. Polynomials are easier to work with as they are composed of terms including powers of the input variable.
The function given in this exercise is \(f(x) = x^3 - 2x^2 + 3x - 1\), which is a polynomial of degree 3. By analyzing such a polynomial, you understand how it changes with its input x.
The first derivative shows where the function is increasing or decreasing. The second derivative provides information on the concavity of the function, and the third derivative gives insight into the abrupt changes in curvature.
In our specific function, we see how the coefficients of the terms affect its overall shape. For polynomial function analysis:
The function given in this exercise is \(f(x) = x^3 - 2x^2 + 3x - 1\), which is a polynomial of degree 3. By analyzing such a polynomial, you understand how it changes with its input x.
The first derivative shows where the function is increasing or decreasing. The second derivative provides information on the concavity of the function, and the third derivative gives insight into the abrupt changes in curvature.
In our specific function, we see how the coefficients of the terms affect its overall shape. For polynomial function analysis:
- The linear term affects the slope.
- The quadratic term affects the curvature or 'bending'.
- The cubic term can change the shape more drastically.
Calculus
Calculus forms the backbone of many techniques we use in approximating functions, with Taylor polynomials being one of them. Calculus involves concepts such as limits, derivatives, integrals, and series expansions.
By leveraging calculus, we're able to analyze and break down functions into simpler polynomial forms. This is executed using Taylor series which is a key topic in calculus. It's through calculus that you acquire the skill to find derivatives and integrals, which directly feed into constructing accurate Taylor polynomials.
In the exercise, we used derivatives up to the third order and constructed a Taylor polynomial that precisely matched the original polynomial function.
A deeper understanding of calculus will enhance your ability to approximate functions with polynomials and make sense of how small changes in input affect the outcome. Thus, studying Taylor polynomials isn't just about polynomial function analysis, but also about strengthening your calculus foundation.
By leveraging calculus, we're able to analyze and break down functions into simpler polynomial forms. This is executed using Taylor series which is a key topic in calculus. It's through calculus that you acquire the skill to find derivatives and integrals, which directly feed into constructing accurate Taylor polynomials.
In the exercise, we used derivatives up to the third order and constructed a Taylor polynomial that precisely matched the original polynomial function.
A deeper understanding of calculus will enhance your ability to approximate functions with polynomials and make sense of how small changes in input affect the outcome. Thus, studying Taylor polynomials isn't just about polynomial function analysis, but also about strengthening your calculus foundation.
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