Question

$\sf Let \: f(x) = {e}^{x + {x}^{2} } \: for \:real \: x. \: From \: among \\ \sf the \: following. \: Choose \: the \: Taylor \: series \\ \sf approximation \: of \: f(x) \: around \: x = 0, \: which \\ \sf includes \: all \: powers \: of \: x \: less \: than \: or \: equal \:to \: 3$

$\sf A. \: 1+x+x²+x³$

$\sf B. \: 1 + x + \frac{3}{2} {x}^{2} + {x}^{3}$
$\sf C. \: 1 + x + \frac{3}{2} {x}^{2} + \frac{7}{6} {x}^{3}$
$\sf D. \: 1 + x + 3 {x}^{2} + 7x {}^{3}$
​

Answer 1

Answer:

$Answer:</p><p>Sum is not convergent .</p><p>Step-by-step explanation:</p><p>First let's take integral part</p><p>\begin{gathered}\int ^{ \infty}_{0} \frac{ {x}^{b} }{( {e}^{ax} )(b!)} \: dx \\ \\ put \: ax = y \: \implies \: dx = \frac{dy}{a} \\ \\ \int ^{ \infty}_{0} \frac{ {( \frac{y}{a}) }^{b} }{( {e}^{y} )(b!)} \: \frac{dy}{a} \\ \\ = \frac{1}{ {a}^{b + 1}.b! } \int ^{ \infty}_{0} {y}^{b} {e}^{ - y} dy \\ \\ using \: gamma \: function \\ \\ = \frac{(b + 1)!}{ {a}^{b + 1}. b!} \\ \\ = \frac{b + 1}{ {a}^{b + 1} } \end{gathered}∫0∞(eax)(b!)xbdxputax=y⟹dx=ady∫0∞(ey)(b!)(ay)bady=ab+1.b!1∫0∞ybe−ydyusinggammafunction=ab+1.b!(b+1)!=ab+1b+1</p><p>Now our question is reduced to</p><p>\displaystyle \sf \sum \limits^{ \infty}_{a = 2} \sum \limits^{ \infty}_{b = 1} \frac{b + 1}{ {a}^{b + 1} }a=2∑∞b=1∑∞ab+1b+1</p><p>Let</p><p>\begin{gathered}let \: z = \frac{1}{a} < 1 \: then \:series \\ \\ s = z + {z}^{2} + ... = \frac{ z }{1 - z} \\ \\ differentiate \: wrt \: z \\ \\ 1 + 2z + 3 {z}^{2} + .... = \frac{(1 - z) + z}{(1 - z) {}^{2} } \\ \\ multiply \: both \: sides \: with \: z \\ \\ z + 2 {z}^{2} + 3 {z}^{3} + .... = \frac{z}{(1 - z) {}^{2} } \\ \\ put \: z = \frac{1}{a} \\ \\ \frac{1}{a} + \frac{2}{ {a}^{2} } + \frac{3}{ {a}^{3} } + .... = \frac{( \frac{1}{a} )}{(1 - \frac{1}{a}) {}^{2} } = \frac{ {a}^{2} }{(a - 1) {}^{2} } \\ \implies \\ \frac{2}{a {}^{2} } + \frac{3}{ {a}^{3} } + .... = \frac{ {a}^{2} }{(a - 1) {}^{2} } - \frac{1}{a} \\ \implies \\ \\ \sum \limits^{ \infty}_{b = 1} \frac{b + 1}{ {a}^{b + 1} } = \frac{a{}^{3} - {a}^{2} + 2a - 1 }{a(a - 1) {}^{2} } \end{gathered}letz=a1<1thenseriess=z+z2+...=1−zzdifferentiatewrtz1+2z+3z2+....=(1−z)2(1−z)+zmultiplybothsideswithzz+2z2+3z3+....=(1−z)2zputz=a1a1+a22+a33+....=(1−a1)2(a1)=(a−1)2a2⟹a22+a33+....=(a−1)2a2−a1⟹b=1∑∞ab+1b+1=a(a−1)2a3−a2+2a</p><p>$