Find the Rates of convergence of the following functions as h->0
lim
h->0 sin h - hcosh
------------- = 0
h
Find the Rates of convergence of the following functions as n->inf
lim (Sin (1/n))^2=0
n-> inf
lim
h->0 (sin h - hcosh)/h
Use L'Hopital's rule. The ratio of derivatives of numerator and denominator is
cosh - h*cosh - h*sinh
and that approaches 1 as h -> 0
im asking the rate of convergence as the order such h^2? h^3? etc
To find the rates of convergence of the functions as h->0 and n->inf, we can use limits and calculus concepts.
For the first function:
lim sin(h) - h*cosh
h->0 ---------------- = 0
h
To find the rate of convergence as h approaches 0, we can use L'Hospital's rule. Differentiating the numerator and denominator with respect to h, we get:
lim cos(h) - cosh - h*sinh
h->0 --------------------- = 0
1
Evaluating the limit, we find that the rate of convergence is 0. This means that the function converges to 0 as h approaches 0.
For the second function:
lim (sin(1/n))^2 = 0
n->inf
As n approaches infinity, (sin(1/n))^2 will converge to 0. In this case, we can reformulate the limit as:
lim sin(1/n)
n->inf
Using the limit of sin(x)/x as x approaches 0, we know that sin(1/n) will converge to 1/n as n approaches infinity. Therefore, the rate of convergence is 1/n.
To find the rate of convergence as a variable approaches zero or infinity, we need to analyze the behavior of the function as the variable gets closer to the limit.
For the first function, as h approaches 0, we can use Taylor series expansion to compute the limit more easily. The Taylor series expansion for sin(h) and cosh(h) are:
sin(h) = h - (h^3)/3! + (h^5)/5! - ...
cosh(h) = 1 + (h^2)/2! + (h^4)/4! + ...
Using this expansion, we can rewrite the function:
lim sin(h) - hcosh(h)
h->0 ---------------
h
= lim (h - h^3/3! + h^5/5! - ...) - h(1 + h^2/2! + h^4/4! + ...)
h->0
------------------------------------
h
= lim h - h^3/3! + h^5/5! - h - h^3/2! - h^5/4! + ...
h->0
--------------------------------------------------
h
= lim -h^3/3! + h^5/5! - h^3/2! - h^5/4! + ...
h->0
-------------------------------------------
h
As h approaches 0, the higher order terms (h^5, h^3) become negligible compared to h. Thus, the leading term in the numerator is -h^3/3!, and the leading term in the denominator is h. Therefore, the rate of convergence of the function as h approaches 0 is h^2.
For the second function, as n approaches infinity, we can analyze the behavior of the function using limits.
lim (sin(1/n))^2
n-> inf
Since sin(x) is bounded between -1 and 1 for any value of x, we know that as n approaches infinity, sin(1/n) approaches 0. Therefore, as n goes to infinity, the function (sin(1/n))^2 also approaches 0.
In this case, the rate of convergence as n approaches infinity is not specified because the function converges to the limit exactly. The function converges to 0 without any specific rate.