Question

Expand log(x+a) in powers of x by Taylor's Theorem.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Given: log(x+a)

To find: expand in the power of x by taylor theorem.

explanation:

according to taylor's theorem, it gives an approximation of a k-times differentiable function around a given point by a polynomial of degree k.

as given to us,

                                log(x+a)

multiply by a in numerator and denominator,

                         = $log(\frac{x+a}{a}).a$

                          = $log(\frac{x}{a} +\frac{a}{a} ).a$

                          = $log$$(\frac{x}{a} +1).a$

as we know that,

                       = log(a×b) = log a + log b

then,

                       = $log$$(1+\frac{x}{a)}+$$log$$a$

                       =$(\frac{x}{a} -\frac{(x/a)^{2} }{2} +\frac{(x/a)^{2} }{3} - ----------+$∞$)+loga$

expansion of log(x+a),

                       =  $loga + \frac{x}{a} -\frac{(x)^{2} }{2a^{2} } +\frac{(x)^{3} }{3a^{3} } - ----------+$∞

hence expansion of loa(x+a) in powers of x by taylor's theorem is,

                      $loga + \frac{x}{a} -\frac{(x)^{2} }{2a^{2} } +\frac{(x)^{3} }{3a^{3} } - ----------+$∞