Given the fact that the Taylor series about x = 0 for $e^{x}=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots$ , is the Taylor series about x = 0 for $e^{x / 4}=1+\frac{x}{4}+\frac{x^{2}}{32}+\frac{x^{3}}{384}+\cdots$ ?

Question

Quizplus · Accepted Answer

The Taylor series for $e^{x/4}$ about $x = 0$ is obtained by substituting $x/4$ for $x$ in the Taylor series of $e^x$, resulting in $e^{x/4} = 1 + \frac{x}{4} + \frac{x^2}{2!(4^2)} + \frac{x^3}{3!(4^3)} + \cdots$, which simplifies to $1 + \frac{x}{4} + \frac{x^2}{32} + \frac{x^3}{384} + \cdots$.

Given the Fact That the Taylor Series About X = ex=1+x1!+x22!+x33!+⋯e^{x}=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdotsex=1+1!x​+2!x2​+3!x3​+⋯

Given the Fact That the Taylor Series About X = $e^{x}=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots$