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uppaladhadium
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We know that log(1+x) = x+((x^2)/2)+((x^3)/3)+....((x^n)/n)+...
Could anybody please tell me the proof
Could anybody please tell me the proof
uppaladhadium said:In taylor series if f(x)=log(1+x) then is f'(a)=0?
and is f(a)=log(1+a)
The expansion of log(1+x) is given by the following formula: log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...
The expansion of log(1+x) is derived using the Maclaurin series, which is a special case of the Taylor series. It involves taking the derivatives of the function at a specific point (in this case, x=0) and plugging them into the formula.
The expansion of log(1+x) is significant because it allows us to approximate the value of log(1+x) for small values of x, which is useful in many mathematical and scientific applications. It also helps in simplifying complex logarithmic expressions and solving certain types of equations.
No, the expansion of log(1+x) is only valid for values of x that result in a convergent series. This means that for the expansion to be valid, the value of x must be within a certain range determined by the domain of the function.
No, the expansion of log(1+x) is an infinite series and can only provide an approximation of the value of log(1+x). To find the exact value, we would need to use a more precise method such as a calculator or a table of logarithms.