10-5. Consider the power series x2 z=x+ 2! x3 + 3! ... nvert this to find x as a power series in z. Compare your result to the expansion of In(1+ z), ince z above is actually e* – 1.
10-5. Consider the power series x2 z=x+ 2! x3 + 3! ... nvert this to find x as a power series in z. Compare your result to the expansion of In(1+ z), ince z above is actually e* – 1.
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![**10-5. Consider the power series**
\[ z = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
Invert this to find \( x \) as a power series in \( z \). Compare your result to the expansion of \( \ln(1+z) \), since \( z \) above is actually \( e^x - 1 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddaece62-c748-41a0-80d0-fafdb1836a11%2F49a1038a-8c24-4b86-bb6e-3a7fdc3ac3f8%2Fr73lay_processed.png&w=3840&q=75)
Transcribed Image Text:**10-5. Consider the power series**
\[ z = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
Invert this to find \( x \) as a power series in \( z \). Compare your result to the expansion of \( \ln(1+z) \), since \( z \) above is actually \( e^x - 1 \).
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