Showing that 2 functions are equal to each other

1. Claire84

221
If you have arcsecx= pi/2 - arccot(sqrt(x^(2) -1) How do you actually show that these are equal to eah other. What we're told at the start of the question is thatonly constant functions have derivative zero at every point of an interval, and using that idea we're to show the relation aobe. Then as a hine we're told to try the values of arcesx and arccot (sqrt(x^(2) -1) when x =sqrt 2

To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively. Can you give me a hint as to where I should be going with this question? Thanks!

2. matt grime

9,395
let f be inverse sec, and g inverese cot,

then you know df/dx=-dg/dx

that is, using that only constant functions have zero derivative,

f=-g+k for some constant k, put in some nice value of x to find what k is

sec(pi/4)= sqrt(2) so pi/4 = arcseec(sqrt(2))

and cot(pi/4)=1 so arccot(1)=pi/4

hence setting x=sqrt2 in the equation f=-g+k says that

pi/4=-pi/4+k, ie k=pi/2

3. HallsofIvy

40,943
Staff Emeritus
"What we're told at the start of the question is that only constant functions have derivative zero at every point of an interval, and using that idea we're to show the relation above."

In other words, if (f-g)'= f'- g'= 0 for all x (or all x on an interval) then f-g is a constant (on that interval). If you can show further that, for one x, say x= a, f(a)- g(a)= 0, then you have shown that that constant is 0: f(x)- g(x)= 0 for all x on the interval so f(x)= g(x) on the interval.

When you say "To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negative respectively", what two functions are you talking about? I think you mean arcsec(x) and arccot(sqrt(x^(2) -1) but the two functions you want to prove equal are arcsec(x) and
pi/2-arccot(sqrt(x^(2) -1). Of course, the derivative of
pi/2-arccot(sqrt(x^(2) -1), because of that "- arccot.." IS precisely the negative of the derivative of arccot(sqrt(x^(2) -1) so you have in fact, shown that the two functions have the same derivative and, so, differ by some constant.

Now if you can show that those two functions have the same value at, say, sqrt(2) (because that's particularly easy!) you will have shown that that constant is 0 and the two functions are equal for all x.

4. Claire84

221
Right, can you just explain the bit about the zero derivative please, like how you can get to f = -g +k ? Likre I understand how you could write that but how do you know this just because of the zero derivative? If we had zero derivative what would happen? The rest of it makes perfect sense btw.

5. Claire84

221
Oh I think I see now how the constant function comes into it, although I'm not sure which method I should be going off. In HoI's method I can see how you write the two functions who's derivative is constant, but for matt's a different approach is taken. Where does the constancy come in there?

6. matt grime

9,395

df/dx=-dg/dx

so rearranging

d?dx(f+g)=0

implies f+g=k some constant k so

f=-g+k

in HoI method it is, I think, that h=pi/2-arccot(junk)

and he shows that df/dx=dh/dx, and that h(sqrt(2))=f(sqrt(2))

I chose mine because you said

"To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively"

so I labelled the functions accordingly, HoI took a different reading of that, that's just where the difference comes from. Nothing special, just that it isn't clear what 'ther two functions' were that you were talking about.

7. HallsofIvy

40,943
Staff Emeritus
That "HoI" stopped me for a moment. I have a friend who, on the internet, uses the charming sobriquet "HogOnIce" and he regularly abbreviates it HOI! It hadn't occured to me that "HallsofIvy" works out the same way.

By the way, matt, I took the "two functions" Clair84 had differentiated to be only arcsec(x) and 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively (rather than pi/2-arccot(sqrt(x^(2) -1)) precisely because, that way, the derivatives are "1 over x sqrt(x^(2) - 1) and the same thing again only the negative respectively". Because of that, arcsec(x) and pi/2-arccot(sqrt(x^(2) -1)) have the same derivatives.

Clair84- "Right, can you just explain the bit about the zero derivative please, like how you can get to f = -g +k ?"

I didn't say anything about that because that is exactly what you "hint" said. It can be proven that "If f'(x)= 0 at every point on an interval then f(x) is a constant" using the mean value theorem.

Recall that the mean value theorem says "If f is continuous on the interval [a,b] and differentiable on (a,b) (in other words f has to be continuous at the two endpoints but not necessarily differentiable there) then there exist c in (a,b) such that
(f(b)- f(a))/(b-a)= f'(c)."

Assume f'(x)= 0 at every point in an interval (so, in particular, it is differentiable at every point). Take a, b to be any two points in that interval and apply the mean value theorm: (f(b)- f(a))/(b-a)= f'(c)= 0 so f(b)- f(a)=0 and f(b)= f(a). Since a, b could be any two points, f is a constant function on the interval.

Now suppose f and g are two functions such that f'(x)= g'(x) at every point of an interval. Then f-g is a differentiable function and (f-g)'= f'-g'= 0. That is, f-g is a constant: f-g= C so f= g+ C.

Last edited: Feb 21, 2004