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Chapter 1: Problem 27

Find a polynomial P of degree \(\leq 5\) with \(\mathrm{P}(0)=1\), \(\mathrm{P}(1)=2, \mathrm{P}^{\prime}(0)=\mathrm{P}^{\prime \prime}(0)=\mathrm{P}^{\prime}(1)=\mathrm{P}^{\prime \prime}(1)=0\).

Short Answer

Expert verified
Question: Find a polynomial P(x) of degree 5 or less such that P(0) = 1, P(1) = 2, P'(0) = 0, P'(1) = 0, P''(0) = 0, and P''(1) = 0. Answer: P(x) = \(1 + C(4x^3 - 3x^4 + x^5)\), where C is any constant.

Step by step solution

01

Write the general form of P(x)

Write a polynomial P(x) of degree 5 or less, allowing for coefficients up to the fifth term (the others will be zero): P(x) = \(a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5\).
02

Find P'(x) and P''(x)

Compute the first and second derivatives of P(x) with respect to x, denoted as P'(x) and P''(x), respectively. Using basic rules of derivation get: P'(x) = \(a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + 5a_5x^4\) P''(x) = \(2a_2 + 6a_3x + 12a_4x^2 + 20a_5x^3\)
03

Apply the given conditions for P(x), P'(x), and P''(x)

Use the given conditions, plugging the appropriate x values into P(x), P'(x), and P''(x), and form a system of linear equations for the coefficients \(a_i\): 1. P(0) = 1, so \(a_0 = 1\) 2. P(1) = 2, so \(a_0 + a_1 + a_2 + a_3 + a_4 + a_5 = 2\) 3. P'(0) = 0, so \(a_1 = 0\) 4. P''(0) = 0, so \(2a_2 = 0 \implies a_2 = 0\) 5. P'(1) = 0, so \(a_1 + 2a_2 + 3a_3 + 4a_4 + 5a_5 = 0\) 6. P''(1) = 0, so \(2a_2 + 6a_3 + 12a_4 + 20a_5 = 0\)
04

Solve the system of linear equations for the coefficients \(a_i\)

From conditions 1-4, we know that \(a_0 = 1\), \(a_1 = 0\), and \(a_2 = 0\). By substituting these values into equations 5 and 6, we are left with a system of linear equations in \(a_3, a_4,\) and \(a_5\): 1. \(3a_3 + 4a_4 + 5a_5 = 0\) 2. \(6a_3 + 12a_4 + 20a_5 = 0\) Divide the second equation by 2 to simplify it: \(a_3 + 2a_4 + 4a_5 = 0\) Now subtract the first equation from the second new equation: \(a_4 + 3a_5 = 0\) From this, we get \(a_4 = -3a_5\). Now substitute this expression for \(a_4\) back into the first equation: \(3a_3 - 12a_5= 0\) Divide by 3: \(a_3 - 4a_5 = 0\) From this, we get that \(a_3 = 4a_5\). Now, we can substitute \(a_3\) and \(a_4\) back into P(x) and simplify:
05

Write the polynomial P(x) using the found coefficients

With the found coefficients substitute their values back into the general form of P(x): P(x) = \(1 + 0 \cdot x + 0 \cdot x^2 + 4a_5x^3 - 3a_5x^4 + a_5x^5\) Simplify the polynomial: P(x) = \(1 + a_5(4x^3 - 3x^4 + x^5)\) Since a_5 could be any constant, we can write the final polynomial with a constant C (C = a_5) as follows: P(x) = \(1 + C(4x^3 - 3x^4 + x^5)\) The polynomial P(x) satisfies all the given conditions and has degree 5, as required.

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Most popular questions from this chapter

Assuming that \(\int\left(\mathrm{e}^{\mathrm{x}} / \mathrm{x}\right) \mathrm{d} \mathrm{x}\) is not elementary (a theorem of Liouville), prove that \(\int 1 / \ln \mathrm{x} \mathrm{dx}\) is not elementary.

Deduce the reduction formula for \(I_{n}=\int \frac{d x}{\left(1+x^{4}\right)^{n}}\) andhenceevaluate \(I_{2}=\int \frac{d x}{\left(1+x^{4}\right)^{2}} .\)

Evaluate the following integrals: (i) \(\int \mathrm{e}^{\mathrm{x}} \frac{\left(\mathrm{x}^{2}-3 \mathrm{x}+3\right)}{(\mathrm{x}+2)^{2}} \mathrm{dx}\) (ii) \(\int \frac{\mathrm{e}^{\mathrm{x}}\left(\mathrm{x}^{2}+1\right)}{(\mathrm{x}+1)^{2}} \mathrm{dx}\) (iii) \(\int \mathrm{e}^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) (iv) \(\int \frac{x^{2} e^{x}}{(x+2)^{2}} d x\)

Evaluate the following integrals : $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} d x $$

Evaluate the following integrals : $$\int \frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}-1}$$
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