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Let $$ p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} $$ Compute the sequence of values \(p(0), p^{\prime}(0), p^{\prime \prime}(0), p^{\prime \prime \prime}(0), \ldots\)

Short Answer

Expert verified
The sequence is \( p(0) = a_0, \ p'(0) = a_1, \ p''(0) = 2a_2, \ p'''(0) = 6a_3, \ldots, \ p^{(k)}(0) = k! \cdot a_k \).

Step by step solution

01

Evaluate p(0)

To find the value of the polynomial \( p(x) \) at \( x = 0 \), substitute 0 for \( x \) in the polynomial. This results in \( p(0) = a_0 + a_1 \cdot 0 + a_2 \cdot 0^2 + \cdots + a_n \cdot 0^n \). Hence, \( p(0) = a_0 \).
02

Differentiate to find p'(x)

Compute the first derivative of the polynomial \( p(x) \). The derivative is \( p'(x) = a_1 + 2a_2 x + 3a_3 x^2 + \cdots + na_n x^{n-1} \).
03

Evaluate p'(0)

Evaluate the first derivative at \( x = 0 \). Substitute 0 into \( p'(x) \) to get \( p'(0) = a_1 + 2a_2 \cdot 0 + 3a_3 \cdot 0^2 + \cdots + na_n \cdot 0^{n-1} \), simplifying to \( p'(0) = a_1 \).
04

Differentiate to find p''(x)

Differentiate \( p'(x) \) to find the second derivative. The result is \( p''(x) = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots + n(n-1)a_n x^{n-2} \).
05

Evaluate p''(0)

Evaluate the second derivative at \( x = 0 \). This gives \( p''(0) = 2a_2 + 6a_3 \cdot 0 + 12a_4 \cdot 0^2 + \cdots = 2a_2 \).
06

Differentiate to find p'''(x)

Compute the third derivative \( p'''(x) \) by differentiating \( p''(x) \). The expression will be \( p'''(x) = 6a_3 + 24a_4 x + 60a_5 x^2 + \cdots \).
07

Evaluate p'''(0)

Evaluate \( p'''(x) \) at \( x = 0 \). This results in \( p'''(0) = 6a_3 \).
08

Establish a pattern

Notice the sequence of expressions \( p(0) = a_0 \), \( p'(0) = a_1 \), \( p''(0) = 2a_2 \), \( p'''(0) = 6a_3 \). The coefficients in these expressions are factorials based on the derivatives. Generally, \( p^{(k)}(0) = k! \cdot a_k \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental process in calculus used to find the rate of change of a function. When dealing with polynomial functions, differentiation involves applying rules to each term of the function to determine its derivative.

A polynomial function like \[ p(x)=a_{0}+a_{1} x+a_{2} x^{2}+ dots+a_{n} x^{n} \] can be differentiated to find not just the function itself, but new functions that describe the rate at which the original is changing. Each of these new functions is a 'derivative.' When differentiating a polynomial term like \(a_k x^k\), the rule is to multiply \(a_k\) by the exponent \(k\) and then reduce the exponent by one. This means in our sequence evaluation, differentiating any term \(a_k x^k\) results in \(k \, a_k \, x^{k-1}\).

This process repeats, as demonstrated in our solution, for each level of derivative—leading to first, second, third derivatives, and so on. Each derivative provides insights into how the function changes at different rates, making differentiation valuable in understanding the behavior of polynomial functions.
Factorials
Factorials are mathematical expressions where a number is multiplied by all the positive integers less than itself down to 1. Factorial notation is expressed as \(k!\), where the \(!\) symbol represents factorial. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\).

In the context of our polynomial exercise, factorials appear naturally within the sequence of derivatives evaluated at zero, such as \(p''(0) = 2a_2\) translating to \(2!a_2\), and \(p'''(0) = 6a_3\) to \(3!a_3\). Factorials help in simplifying and understanding the pattern in derivatives, especially when evaluating them at a point such as zero.
  • They show how the coefficients of the derivatives change systematically with each order.
  • It underlies the predictive pattern in the evaluated sequence, allowing one to easily generalize results.
Overall, recognizing factorials' role in this context helps pave the way for more complex calculus problems and understanding.
Sequence Evaluation
Sequence evaluation in this exercise involves calculating a sequence of values, specifically \( p(0), p'(0), p''(0), p'''(0), \) and so forth, by evaluating the function and its derivatives at a certain point—in this case, \(x=0\).

For the given polynomial, the sequence evaluation begins by determining what the polynomial gives directly when \(x\) is zero. Following this, each derivative is found and then also evaluated at zero. This involves substituting zero for \(x\) in each derived function.
  • Begin with the original function to find \(p(0)\), in which all terms with \(x\) reduce, leaving just \(a_0\).
  • Apply differentiation successively, simplifying each result at \(x=0\).
  • Notice a pattern: the evaluated results align with factorial changes in coefficient importance.
The exercise of sequence evaluation thus highlights the strategic approach to understanding polynomial function behaviors through systematic calculations, shedding light on deeper properties such as linearity, curvature, and beyond.

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Most popular questions from this chapter

Interpreting the slope of a chord as an average rate of change and the derivative as an instantaneous rate of change, what does the mean value theorem say? If a car travels 100 miles in 2 hours, and the position \(s(t)\) of the car at time \(t\) satisfies the hypotheses of the mean value theorem, can we be sure that there is at least one instant at which the velocity is \(50 \mathrm{mph}\) ?

Let \(A=\left\\{a_{1}, a_{2}, a_{3}, \ldots\right\\}\) be any countable set of real numbers. Let $$ f(x)=\sum_{1}^{\infty} \frac{\left|x-a_{k}\right|}{10^{k}} $$ Prove that \(f\) is convex on \(\mathbb{R}\), differentiable on the set \(\mathbb{R} \backslash A\), and nondifferentiable on the set \(A\).

(Inflection Points) In elementary calculus one studies inflection points. The definitions one finds try to capture the idea that at such a point the sense of concavity changes from strict "up to down" or vice versa. Here are three common definitions that apply to differentiable functions. In each case \(f\) is defined on an open interval \((a, b)\) containing the point \(x_{0}\). The point \(x_{0}\) is an inflection point for \(f\) if there exists an open interval \(I \subset(a, b)\) such that on \(I\) (Definition A) \(f^{\prime}\) increases on one side of \(x_{0}\) and decreases on the other side. (Definition B) \(f^{\prime}\) attains a strict maximum or minimum at \(x_{0}\). (Definition C) The tangent line to the graph of \(f\) at \(\left(x_{0}, f\left(x_{0}\right)\right)\) lies below the graph of \(f\) on one side of \(x_{0}\) and above on the other side. (a) Prove that if \(f\) satisfies Definition \(\mathrm{A}\) at \(x_{0}\), then it satisfies Definition \(\mathrm{B}\) at \(x_{0}\) (b) Prove that if \(f\) satisfies Definition \(\mathrm{B}\) at \(x_{0}\), then it satisfies Definition \(\mathrm{C}\) at \(x_{0}\) (c) Give an example of a function satisfying Definition \(\mathrm{B}\) at \(x_{0}\), but not satisfying Definition A. (d) Give an example of an infinitely differentiable function satisfying Definition \(\mathrm{C}\) at \(x_{0}\), but not satisfying Definition \(\mathrm{B}\). (e) Which of the three definitions states that the sense of concavity of \(f\) is "up" on side of \(x_{0}\) and "down" on the other?

Find an example of an everywhere differentiable function \(f\) so that \(f^{\prime}\) is not everywhere continuous.

If \(f\) and \(g\) are convex on an interval \(I\), show that any linear combination \(\alpha f+\beta g\) is also convex provided \(\alpha\) and \(\beta\) are nonnegative.

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