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 | 1 | initial version |
Given the example, it is pretty easy:
sage: K=var('K');
....: l=var('l');
....: x=var('x');
....: z(x)=(cosh(-1/2*K*l+K*x)-cosh(1/2*K*l))/K;
The derivative of z (with respect to the variable x) is:
sage: diff(z(x),x)
sinh(-1/2*K*l + K*x)
Since the only zeros of the sinh function is zero, solving $z'(x)=0$ is equivalent to solving $-1/2Kl + K*x=0$:
sage: solve(-1/2*K*l + K*x == 0, x)
[x == 1/2*l]
So, if there is an extrema, it can only be at $x=l/2$, but it is not sure.
However, since sinh changes its sign at 0, and since -1/2*K*l + K*x is affine, z' changes its sign at l/2 (when K is nonzero), hence the point $x=l/2$ is an extrema (hence, the only one). The nature of the extrema depends on the sign of K.
You can check with different values of K and l, e.g.:
sage: z.subs(K=2,l=2).plot([-1,2])
sage: z.subs(K=-2,l=2).plot([-1,2])
sage: z.subs(K=1,l=3).plot([-1,3])
sage: z.subs(K=-1,l=3).plot([-1,3])
| | 2 | No.2 Revision |
Given the example, it is pretty easy:
sage: K=var('K');
....: l=var('l');
....: x=var('x');
....: z(x)=(cosh(-1/2*K*l+K*x)-cosh(1/2*K*l))/K;
The derivative of z (with respect to the variable x) is:
sage: diff(z(x),x)
sinh(-1/2*K*l + K*x)
Since the only zeros of the sinh function is zero, solving $z'(x)=0$ is equivalent to solving $-1/2Kl + K*x=0$:
sage: solve(-1/2*K*l + K*x == 0, x)
[x == 1/2*l]
EDIT Actually Sage knows about sinh:
sage: solve(diff(z(x),x)==0,x)
[x == 1/2*l]
So, if there is an extrema, it can only be at $x=l/2$, but it is not sure.
However, sure yet.
Now, since sinh changes its sign at 0, and since -1/2*K*l + K*x is affine, z' changes its sign at l/2 (when K is nonzero), hence the point $x=l/2$ is an extrema (hence, the only one). The nature of the extrema depends on the sign of K.
You can check with different values of K and l, e.g.:
sage: z.subs(K=2,l=2).plot([-1,2])
sage: z.subs(K=-2,l=2).plot([-1,2])
sage: z.subs(K=1,l=3).plot([-1,3])
sage: z.subs(K=-1,l=3).plot([-1,3])
