 Sahil Khan
1

A Maclaurin series is a power series that allows one to calculate an approximation of a function f(x) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero.

We know the formula for a Taylor series centered at x = 0 is

f(0)+f'(0)x+ \frac{f^{n}(0)}{2!}x^{2}+ \frac{f^{n}(0)}{3!}x^{3}+.....

The given function is f(x) = \sum_{n=0}^{\infty}\Big(\frac{-x}{4}\Big)^{n}

f(x)= 1 - \frac{x}{4} + \frac{x^{2}}{4^{2}} - \frac{x^{3}}{4^{3}}

f(3) = 1 - \frac{3}{4} + \frac{9}{16} - \frac{27}{64}

S = \frac{1}{1 - (\frac{-3}{4})} = \frac{1}{\frac{7}{4}} = \frac{4}{7}

So, the value of f(3) = \frac{4}{7} Qalaxia QA Bot
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I found an answer from mathematica.stackexchange.com

Series approximation to integral - Mathematica Stack Exchange

g = Integrate[ Exp[-(x - y)^2/(2 y s^2)] f[y] 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] ... the expansion sg1=xk+σ2k(1+k)2xk−1 can be related to the Taylor expansion of ... (* {1/(1 + x), (-s^2 + (1 + x)^2)/(1 + x)^3, ( 3 s^4 - s^2 (1 + x)^2 + (1 + x)^4)/(1 + x)^5} *) ... ImageSize -> 400, PlotLabel -> Style["s-expansion of integral\nf(y) = 1/(1+y),  ...

For more information, see Series approximation to integral - Mathematica Stack Exchange Qalaxia Knowlege Bot
0

I found an answer from web.stanford.edu

the solutions to Exam 1, blue version

a) (4 points) Give the definition of the Taylor polynomial of degree N of the function f. ... of degree 3 centered at a = 0 of the function f:(-1/2, 1/2) + R given by the rule ... converges when x = -4 and diverges when 2 = 8. ... So IRN (O.) 1 1:(0.1 ) NF.