leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

[Paradoxica]

Yes. But take care with signs.

 

[leehuan]

Yes. But take care with signs.

Right. From memory it's the derivatives of the reciprocals (of the hyperbolic functions) that get the negative?

As opposed to trig where it was the co-functions

 

[Paradoxica]

Right. From memory it's the derivatives of the reciprocals (of the hyperbolic functions) that get the negative?

As opposed to trig where it was the co-functions

actually it would be safer to use the exponential definitions...

 

[leehuan]

Oh actually that's clever

 

[seanieg89]

Just remember that sinh and cosh don't change sign when differentiated, and deduce all sign things from that. I don't memorise this stuff despite using it pretty often.

 

[InteGrand]

For the csch it works the same, because when we multiply top and bottom by csch(x) + coth(x) (which is analogous to what we'd do to integrate csc(x) in the circular trig. cases), we see that the derivatives of csch(x) and coth(x) match their circular trig. analogs. I.e. (coth(x))′ = -csch²(x) and (csch(x))′ = -csch(x)*coth(x).

However, if we try doing the same thing for sech(x) as we'd do for the sec(x) (i.e. multiply top and bottom by sech(x) + tanh(x)), we see that it doesn't quite work out. This is because the derivative of sech(x) is not sech(x)*tanh(x), but rather -sech(x)*tanh(x) (a negative sign pops out). For this reason, the antiderivative of sech(x) is of a different shape to that of sec(x). To find the antiderivative of sech(x), use either the exponential definitions or write it as 1/cosh(x) and multiply top and bottom by cosh(x), and then use the identity cosh²(x) = 1 + sinh²(x) on the denominator. This will lead to an answer of tan^-1(sinh(x)).

 

[leehuan]

Yeah I know, it's convenient to use just sinh and cosh.

But I think I just noticed the pattern the day I learnt them and went with the flow

 

[leehuan]

For the csch it works the same, because when we multiply top and bottom by csch(x) + coth(x) (which is analogous to what we'd do to integrate csc(x) in the circular trig. cases), we see that the derivatives of csch(x) and coth(x) match their circular trig. analogs. I.e. (coth(x))′ = -csch²(x) and (csch(x))′ = -csch(x)*coth(x).

However, if we try doing the same thing for sech(x) as we'd do for the sec(x) (i.e. multiply top and bottom by sech(x) + tanh(x)), we see that it doesn't quite work out. This is because the derivative of sech(x) is not sech(x)*tanh(x), but rather -sech(x)*tanh(x). For this reason, the antiderivative of sech(x) is of a different shape to that of sec(x). To find the antiderivative of sech(x), use either the exponential definitions or write it as 1/cosh(x) and multiply top and bottom by cosh(x), and then use the identity cosh₂(x) = 1 + sinh²(x) on the denominator.

Ohhh point

 

[InteGrand]

Right. From memory it's the derivatives of the reciprocals (of the hyperbolic functions) that get the negative?

As opposed to trig where it was the co-functions

Pretty much all of the hyperbolic trig. ones are the same as their circular trig. counterparts in terms of derivatives. The only one that changes is that sech(x) gets a negative sign when differentiated (and of course that cosh(x) doesn't get a negative when differentiated; this is essentially the cause of the sign change for sech(x)'s one). You can see a table of the derivatives of the hyperbolic functions in the blue section near the bottom of this page: http://tutorial.math.lamar.edu/Classes/CalcI/DiffHyperTrigFcns.aspx .

 

[Paradoxica]

Oh actually that's clever

um...

 

[leehuan]

um...

At first I misinterpreted for log by accident then I realised wait it's inverse tangent. So I forgot to change the thing in the brackets.

 

[leehuan]

Having a mind blank.

Would you say y=x/(9-x^2) is not continuous at x=3, x=-3 but still monotonic increasing?

W.A. for graph for convenience http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=x/(x^2-9)

 

[Paradoxica]

Having a mind blank.

Would you say y=x/(9-x^2) is not continuous at x=3, x=-3 but still monotonic increasing?

W.A. for graph for convenience http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=x/(x^2-9)

It would be monotonic in any of the three restricted domains, but not in the entire domain.

 

[KingOfActing]

Having a mind blank.

Would you say y=x/(9-x^2) is not continuous at x=3, x=-3 but still monotonic increasing?

W.A. for graph for convenience http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=x/(x^2-9)

I'd say it's strictly increasing on each of the sub intervals of it's domain that have no discontinuities. x > y implies f(x) > f(y) only if x,y <-3 or -3 < x,y < 3 or 3 < x,y

 

[leehuan]

Knew I was missing something. Thanks.

 

[Paradoxica]

I'd say it's strictly increasing on each of the sub intervals of it's domain that have no discontinuities. x > y implies f(x) > f(y) only if x,y <-3 or -3 < x,y < 3 or 3 < x,y

Isn't that the definition of monotonically increasing or decreasing?

f(x)>f(y) <=> x>y

Reverse the direction for the other one.

But yeah, I'm pretty sure this is the definition of monotonically increasing, which means there cannot be any discontinuities that jump downwards.

This does leave room for upward jumping discontinuities, but that's not a big concern.

 

[leehuan]

So far with proving that a function is injective I know of two methods:

1. f(x)=f(y) <=> x=y
2. monotone

Is there anything else that might interest me?

Also, this would not be considered surjective right:
f: [0,inf) -> R, f(x)=x^2

 

[KingOfActing]

Isn't that the definition of monotonically increasing or decreasing?

f(x)>f(y) <=> x>y

Reverse the direction for the other one.

But yeah, I'm pretty sure this is the definition of monotonically increasing, which means there cannot be any discontinuities that jump downwards.

This does leave room for upward jumping discontinuities, but that's not a big concern.

Strictly increasing. Monotonic increasing is <=, strictly is <.

 

[InteGrand]

So far with proving that a function is injective I know of two methods:

1. f(x)=f(y) <=> x=y
2. monotone

Is there anything else that might interest me?

Also, this would not be considered surjective right:
f: [0,inf) -> R, f(x)=x^2

Correct, not surjective.

 

[Paradoxica]

So far with proving that a function is injective I know of two methods:

1. f(x)=f(y) <=> x=y
2. monotone

Is there anything else that might interest me?

Also, this would not be considered surjective right:
f: [0,inf) -> R, f(x)=x^2

It would be surjective if the codomain was restricted, but given it isn't, it's not surjective.