Starting little-oh notation (2 Viewers)

[leehuan]

LaTeX is broken again so I won't use it

Am I doing this correctly?

Q: As h->0, show that (explain why)
sin(x+h)=sinx+hcosx+o(h)

A: Let g(x)=sin(x+h)-sinx-hcosx

Need |g(x)|<=K|h| for all K>0
i.e. |sin(x+h)-sinx-hcosx|<=K|h|
<=> | (sin(x+h)-sinx)/h - cosx| <= K (as h neq 0)
LHS = | (sin(x+h)-sinx)/h - cosx| = | cosx - cosx | as by limiting h to 0 we can use the definition of the derivative = 0 <= K trivially

Hence, |g(x)|<=K|h| as required
<=> g(x) = o(h)
<=> sin(x+h)=sinx+hcosx+o(h) QED

 

[InteGrand]

LaTeX is broken again so I won't use it

Am I doing this correctly?

Q: As h->0, show that (explain why)
sin(x+h)=sinx+hcosx+o(h)

A: Let g(x)=sin(x+h)-sinx-hcosx

Need |g(x)|<=K|h| for all K>0
i.e. |sin(x+h)-sinx-hcosx|<=K|h|
<=> | (sin(x+h)-sinx)/h - cosx| <= K (as h neq 0)
LHS = | (sin(x+h)-sinx)/h - cosx| = | cosx - cosx | as by limiting h to 0 we can use the definition of the derivative = 0 <= K trivially

Hence, |g(x)|<=K|h| as required
<=> g(x) = o(h)
<=> sin(x+h)=sinx+hcosx+o(h) QED

Well what they're asking you to show is essentially the fact that sin(x) has derivative cos(x). This is because a function f being differentiable at a satisfies f(a+h) = f(a) + h*f'(a) + o(h) (as h -> 0).

Conversely, if a function satisfies f(a+h) = f(a) + h*A + o(h), where A is independent of h (can depend on a though), then f is differentiable at a with derivative there A = A(a) = f'(a).

(I think seanieg89 posted a general Q about this in the 4U marathon this year, so you can try finding that.)

Since you're allowed to just 'explain' why that is true, yes, you should be allowed to do it by differentiation (basically it's sufficient to say that f is differentiable everywhere with derivative f'(x) = cos(x)).

Btw, note that an equivalent definition of f(x) = o(g(x)) (as x -> a) is that f(x)/g(x) -> 0 as x -> a.

 

[leehuan]

Oh right, thanks for the equivalency. Completely forgot about that

Another question.

One way to do this is by the squeeze theorem. But will properties of limits also work?

lim h->0 2hsin(1/h) = (lim h->0 2h).(lim h->0 sin(1/h)) using properties of limits
= 0 . 0 keeping in mind that sin is continuous at 0 and of course lim z->0 1/z = 0
= 0

 

[InteGrand]

Oh right, thanks for the equivalency. Completely forgot about that

Another question.

One way to do this is by the squeeze theorem. But will properties of limits also work?

lim h->0 2hsin(1/h) = (lim h->0 2h).(lim h->0 sin(1/h)) using properties of limits
= 0 . 0 keeping in mind that sin is continuous at 0 and of course lim z->0 1/z = 0
= 0

No, limit as h -> 0 of sin(1/h) does not exist. Best to use squeeze law. Limit as z -> 0 of 1/z isn't 0, but rather it does not exist. (1/z approaches +inf as x -> 0 from above and -inf as as x -> 0 from below.)

 

[leehuan]

Clumsiness again. Ok thanks squeeze theorem it is

 

[leehuan]

How would you properly word your answer to part b)?





Qn:




(Basically, I don't know how to properly word/rephrase/manipulate the definition |f(x)| < K|g(x)|

 

[InteGrand]

How would you properly word your answer to part b)?





Qn:




(Basically, I don't know how to properly word/rephrase/manipulate the definition |f(x)| < K|g(x)|

By definition of limits, since f(x)/g(x) -> 0 as x -> oo, we know for ANY eps > 0, we have

|f(x)/g(x) - 0| < eps <=> f(x)/g(x) < eps, for all x greater than some M (i.e. for all x sufficiently large).

In particular, we can choose eps = 1 and get that f(x)/g(x) < 1 <=> f(x) < g(x) for all x sufficiently large.

(Oh, here I actually used the definition of f(x) = o(g(x)) meaning that f(x)/g(x) -> 0, but we got your K definition out of this from definition of limits. If you wanted to start at your K definition, note that it says that for ANY K > 0, we have |f(x)| < K |g(x)| for all x sufficiently large. So since it's the case for ANY positive K, it is in particular the case for K = 1, which gives us the desired inequality. Of course the absolute value signs can be ignored here since the functions in question are positive.)

 

[leehuan]

How would you smack out the absolute values here? Because it's not intuitive to me that sinh(x^-1) - x^-1 > 0

 

[InteGrand]

How would you smack out the absolute values here? Because it's not intuitive to me that sinh(x^-1) - x^-1 > 0

(Also you can easily see it if you remember the graph of sinh t. It has the line y = t as a tangent at the origin, and is convex ("concave up") on the positive t-axis, so will be above the line always after t = 0.)

Also I think you typoed the Q, because you said to explain the little-oh thing but you said that was already proved. Was it meant to be explain why it's < x^-2 instead?

 

[leehuan]

(Also you can easily see it if you remember the graph of sinh t. It has the line y = t as a tangent at the origin, and is convex ("concave up") on the positive t-axis, so will be above the line always after t = 0.)

Also I think you typoed the Q, because you said to explain the little-oh thing but you said that was already proved. Was it meant to be explain why it's < x^-2 instead?

Yep. That exact typo. lol