# 2 < e < 3

1. Nov 2, 2005

### Karlsen

I have to carry out a few steps to show that 2 < e < 3. (e as in 2.71828..)

Let f(t) = 1/t for t > 0.

(a)
Show that the area under y = f(t), above y = 0, and between t = 1 and t = 2 is less than 1 square unit. Deduce that e < 2.

This is easy, just integrating and getting ln(2). From graph I see that ln(2) < 1 => e > 2.

(b)
Show that all tangent lines to the graph of f lie below the graph.

Also not very hard; just showing that f'(t) < 0 gives that it's a decreasing function in the interval ]0,inf[, so the tangent lines have negative slopes, hence lies below the graph.

(c)
Find the lines T_2 and T_3 that are tangent to y = f(t) at t = 2 and t = 3, respectively.

I got
T_2: y = (-1/4)t + 1
T_3: y = (-1/9)t + (2/3)

(d)
Find the area A_2 under T_2, above y = 0, and between t = 1 and t = 2. Also find the area A_3 under T_3, above y = 0, and between t = 2 and t = 3.

I got
A_2 = 5/8
A_3 = 7/18

(e)
Show that A_2 + A_3 > 1 square unit. Deduce that e < 3.

OK, so A_2 + A_3 = (5/8) + (7/18) = 73/72 > 1, but I can't see how this leads to e < 3. Any tips or suggestions?

2. Nov 2, 2005

### Karlsen

Oh, sorry. I guess I Didn't think long enough.

int(t=1,t=3, 1/t) > A_2 + A_3 > 1, so ln(3) > 1, hence e < 3. bam!

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