求过原点的y=tanh(x)曲线切线方程及cx=tanh(x)临界c值

Alright, let's work through this problem step by step—first finding the tangent lines to y = tanh(x) that pass through the origin, then using that to figure out the critical value of positive constant c for which cx = tanh(x) has non-trivial (i.e., non-zero) solutions.

First, recall that the derivative of tanh(x) is sech²(x) = 1 - tanh²(x)—this gives the slope of the tangent line at any point x.

Tangent at the origin #

The curve y = tanh(x) passes through the origin (tanh(0) = 0). The slope here is tanh'(0) = sech²(0) = 1, so the tangent line at the origin is straightforward:

y = x

Are there other tangents through the origin? #

Suppose there's a tangent line that passes through the origin and touches the curve at some point (a, tanh(a)) where a ≠ 0. The tangent line equation using point-slope form is:

y - tanh(a) = (1 - tanh²(a))(x - a)

Since this line goes through (0, 0), substitute x=0 and y=0 into the equation:

-tanh(a) = (1 - tanh²(a))(-a)
tanh(a) = a(1 - tanh²(a))

Let's rewrite this using hyperbolic identities to simplify:

• Multiply both sides by cosh²(a) (which is always positive):

  tanh(a)cosh²(a) = a(1 - tanh²(a))cosh²(a)
  

• tanh(a)cosh²(a) = sinh(a)cosh(a) = (sinh(2a))/2
• (1 - tanh²(a))cosh²(a) = cosh²(a) - sinh²(a) = 1

So the equation simplifies to:

sinh(2a) = 2a

Now consider the function g(z) = sinh(z) - z for real z. Its derivative is g'(z) = cosh(z) - 1, which is non-negative for all z (since cosh(z) ≥ 1), and equals 0 only at z=0. This means g(z) is strictly increasing for z > 0, and g(0) = 0. So sinh(z) > z for all z > 0, and sinh(z) < z for all z < 0.

This tells us the equation sinh(2a) = 2a has no non-zero solutions. The only tangent line to y = tanh(x) that passes through the origin is y = x.

Now let's use this tangent line to analyze when cx = tanh(x) has solutions other than x=0:

• When c > 1: For x > 0, we know tanh(x) < x (since tanh(x) is concave down for x > 0, and its tangent at the origin is y=x). Since c > 1, cx > x > tanh(x) for all x > 0. For x < 0, tanh(x) > x (odd function property), and cx < x (since c > 1 and x is negative), so tanh(x) > x > cx. The only solution here is x=0.
• When c = 1: The equation becomes x = tanh(x). As mentioned, tanh(x) ≤ x for all real x, with equality only at x=0. Again, only the trivial solution exists.
• When 0 < c < 1: For small x > 0, tanh(x) ≈ x - x³/3, so tanh(x) - cx ≈ x(1 - c) - x³/3 > 0 (since 1 - c > 0 and x is small). As x → +∞, tanh(x) → 1 while cx → +∞, so tanh(x) - cx → -∞. By the Intermediate Value Theorem, there must be some positive x where tanh(x) = cx. By the oddness of both tanh(x) and cx, there's also a corresponding negative solution. So non-trivial solutions exist here.

Conclusion #

The critical value of positive constant c is 1:

• If c < 1, cx = tanh(x) has two non-trivial solutions (one positive, one negative).
• If c ≥ 1, only the trivial solution x=0 exists.

内容的提问来源于stack exchange，提问作者user534363