The Taylor series expansion of f(x) about x=0 is f(x)=f(0)+f''(0)*x+1/2!*f''(0)*x^2+...., so provided x is near zero, we can truncate the series after the linear term (this is because x^2 will be negligible compared to x if x is very small). So that gives f(x)~f(0)+f'(0)*x. For the function f(x)=(1+x)^n, we have f(0)=1 and f'(x)=n(1+x)^(n-1) which means f'(0)=n, so the truncated Taylor series is f(x)~1+nx. So in the end, you can remember it's just the exponent coming down as the coefficient of x, but personally I go through and scratch out the derivation more often than not -- because I don't use the binomial approximation often enough to remember it with confidence! I made a video on Taylor series generally a few years ago, not sure I ever made one about the binomial approximation in particular. ua-cam.com/video/FNwWn1co7Wc/v-deo.html
So for Taylor series method, I just make the exponent to the coefficient?
The Taylor series expansion of f(x) about x=0 is f(x)=f(0)+f''(0)*x+1/2!*f''(0)*x^2+...., so provided x is near zero, we can truncate the series after the linear term (this is because x^2 will be negligible compared to x if x is very small). So that gives f(x)~f(0)+f'(0)*x. For the function f(x)=(1+x)^n, we have f(0)=1 and f'(x)=n(1+x)^(n-1) which means f'(0)=n, so the truncated Taylor series is f(x)~1+nx. So in the end, you can remember it's just the exponent coming down as the coefficient of x, but personally I go through and scratch out the derivation more often than not -- because I don't use the binomial approximation often enough to remember it with confidence!
I made a video on Taylor series generally a few years ago, not sure I ever made one about the binomial approximation in particular. ua-cam.com/video/FNwWn1co7Wc/v-deo.html