EinstenICEBERG
Einstein V2
- Joined
- Apr 7, 2011
- Messages
- 220
- Gender
- Male
- HSC
- 2014
Before we move on to more rules for the derivative, we’ll introduce a useful
application of it known as first-order Taylor approximations, according to which
we can approximate the change in the value of a function by using its derivative.2
Specifically, we have
f(x + h) = f(x) + (df(x)/dx)*h + R(x, h)h
where R(x, h) is some average approximation error that will approach zero as h
approaches zero. We won’t prove this theorem, but understand it using figure
(3). Recall that when taking a derivative, we are calculating the slope of a
function in one point, represented by the slope of the tangent line in the figure.
The Taylor approximation works by assuming that the slope of the function
is constant and approximating f(x + h) by the value of the tangent at x + h.
Since the derivative of f(x) is not generally constant, this gives rise to the
(total) approximation error R(x, h)h. However, as we let h approach zero, we
are approaching the point where the slope of the function is indeed exactly
df(x)/dx, and R(x, h) will also approach zero.
can someone simplify this for me, I trried understanding through many links but I don't get it still =(
application of it known as first-order Taylor approximations, according to which
we can approximate the change in the value of a function by using its derivative.2
Specifically, we have
f(x + h) = f(x) + (df(x)/dx)*h + R(x, h)h
where R(x, h) is some average approximation error that will approach zero as h
approaches zero. We won’t prove this theorem, but understand it using figure
(3). Recall that when taking a derivative, we are calculating the slope of a
function in one point, represented by the slope of the tangent line in the figure.
The Taylor approximation works by assuming that the slope of the function
is constant and approximating f(x + h) by the value of the tangent at x + h.
Since the derivative of f(x) is not generally constant, this gives rise to the
(total) approximation error R(x, h)h. However, as we let h approach zero, we
are approaching the point where the slope of the function is indeed exactly
df(x)/dx, and R(x, h) will also approach zero.
can someone simplify this for me, I trried understanding through many links but I don't get it still =(