and Approximations 3.5 A Large-T Approximation So far, weve focused on the problem of approximating a function f(r)when is small But what if we need to approximate a function f(ar)when r is large? Here's a simple trick that often works. As an example, let's take our old friend f(ar) In(1 +a). When a is a big number, there is little difference between In x and In(1+r).S let's first write In(1+a) in terms of In x In(1+a)=In In z+In[1+ The general idea is to express f(a) in terms of a fairly similar function and then use Tay- lor's Theorem just to analyze the leftover bit which is In(1+1/a) in this case. notice that when a is large, 1/ is small. So our earlier methods should now give excellent results Specifically, suppose we set z=1/ and use the bounds that we derived earlier ≤ln(1+2)≤ Then we get the upper bound 1(1+x)≤ln and the lower bound: +1/x For example, suppose we wanted to evaluate In(e+1). Then these bounds say e10+1 ≤mn(e0+1)≤10 The difference between the upper and lower bounds is only about e. Not bad!� � �� � � Sums and Approximations 15 3.5 A Large­x Approximation So far, we’ve focused on the problem of approximating a function f(x) when x is small. But what if we need to approximate a function f(x) when x is large? Here’s a simple trick that often works. As an example, let’s take our old friend f(x) = ln(1 + x). When x is a big number, there is little difference between ln x and ln(1 + x). So let’s first write ln(1 + x) in terms of ln x: 1 ln(1 + x) = ln x · 1 + x 1 = ln x + ln 1 + x The general idea is to express f(x) in terms of a fairly similar function and then use Tay￾lor’s Theorem just to analyze the leftover bit, which is ln(1 + 1/x) in this case. Notice that when x is large, 1/x is small. So our earlier methods should now give excellent results. Specifically, suppose we set z = 1/x and use the bounds that we derived earlier: z z ≤ ln(1 + z) ≤ z 1 + Then we get the upper bound 1 ln(1 + x) ≤ ln x + x and the lower bound: 1/x ln(1 + x) ≥ ln x + 1 + 1/x 1 ≥ ln x + x + 1 For example, suppose we wanted to evaluate ln(e10 + 1). Then these bounds say: 1 1 10 + e10 + 1 ≤ ln(e 10 + 1) ≤ 10 + e10 The difference between the upper and lower bounds is only about e−20. Not bad!