Questions: Opgave 2.5 For funktionen f(x)=5 e^x ln (x), bestem femtegrads Taylorpolynomiet for f med udviklingspunkt a=1. Det er følgende funktion: P5(x)=5 e(x-1)+5/2 e(x-1)^2+5/3 e(x-1)^3+3/square e(x-1)^5 . Dit svar skal være et helt tal mellem 0 og 99.

Solution Steps

We start with the function \( f(x) = 5 e^x \ln(x) \).

We compute the first five derivatives of \( f(x) \):

1. \( f'(x) = 5 e^x \ln(x) + 5 e^x \cdot \frac{1}{x} \)
2. \( f''(x) = 5 e^x \ln(x) + 10 e^x \cdot \frac{1}{x} - 5 e^x \cdot \frac{1}{x^2} \)
3. \( f'''(x) = 5 e^x \ln(x) + 15 e^x \cdot \frac{1}{x} - 15 e^x \cdot \frac{1}{x^2} + 10 e^x \cdot \frac{1}{x^3} \)
4. \( f^{(4)}(x) = 5 e^x \ln(x) + 20 e^x \cdot \frac{1}{x} - 30 e^x \cdot \frac{1}{x^2} + 40 e^x \cdot \frac{1}{x^3} - 30 e^x \cdot \frac{1}{x^4} \)
5. \( f^{(5)}(x) = 5 e^x \ln(x) + 25 e^x \cdot \frac{1}{x} - 50 e^x \cdot \frac{1}{x^2} + 100 e^x \cdot \frac{1}{x^3} - 150 e^x \cdot \frac{1}{x^4} + 120 e^x \cdot \frac{1}{x^5} \)

Next, we evaluate each derivative at \( x = 1 \):

• \( f(1) = 0 \)
• \( f'(1) = 13.5914091422952 \)
• \( f''(1) = 13.5914091422952 \)
• \( f'''(1) = 27.1828182845905 \)
• \( f^{(4)}(1) = 0 \)
• \( f^{(5)}(1) = 122.322682280657 \)

The missing coefficient for the \( (x-1)^5 \) term in the Taylor polynomial is given by: \[ \text{Missing Coefficient} = \frac{f^{(5)}(1)}{5!} = \frac{122.322682280657}{120} = 1.01935568567214 \]

This coefficient is the value that needs to be filled in the polynomial for the \( (x-1)^5 \) term.

Final Answer