Question

Q.4. Expand log x in the power of (x-1) and hence evaluate
Expand close to the
log1.2 correct to four decimal places​

Answer 1

Answer:

I hope help for you..........

Answer 2

Answer:

log(1.2) = 0.2062

Step-by-step explanation:

Let f (x) = log x

f'(x) = $\frac{1}{ x^{} }$

f"(x) = -$\frac{1}{ x^{2} }$

f"'(x) = $\frac{2}{ x^{3} }$

Now, we know that Taylor series expansion of a function f(x) about a number 'a' is :

f(x) = Σ $\frac{f^{(n)}(a) . (x-a)^{n} }{n!}$

So, Taylor series expansion of a function log x  about 1 is

=> f(x) =  f (1) . (x - 1 )⁰ + f' (1) .(x - 1 )¹  +  f" (1) . $\frac{(x-1)^{2} }{2!}$ + . .

Now substituting the values of f(x), f'(x) , f''(x), . .we get

f(x) = 0 + (x -1) - $\frac{(x-1)^{2} }{2}$ +  $\frac{(x-1)^{3} }{3}$  -  $\frac{(x-1)^{4} }{4}$  + . .

=>f(1.2) = 0 + (1.2 -1) - $\frac{(1.2-1)^{2} }{2}$ + $\frac{(1.2-1)^{3} }{3}$ - $\frac{(1.2-1)^{4} }{4}$ + . .

=>f(1.2) = 0 + 0.2 - 0.02 + 0.0266 - 0.0004 + . .

=>f(1.2) = 0.2062

=> log(1.2) = 0.2062