Function equal up to nth order, diffbility

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In summary: For example, you might say that if the two functions are equal to the first order at x= a then the limit exists. Then your proof is valid.
  • #1
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Homework Statement


Two functions f,g:R->R are equal up to nth order at if

lim h-->0 [f(a+h) - g(a+h) / hn ]= 0

Show that f is differentiable at a iff there is a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.

Homework Equations



f is differentiable at a if the following limit exists:

lim h-->0 [f(a+h) - f(a) / h]


The Attempt at a Solution



Direction 1:
=>

Assume f is differentiable. Show there exists a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.

So, lim h-->0 [f(a+h) - f(a) / h] exists and equals some L.

=>lim h-->0 [f(a+h) - f(a) - g(a+h) + g(a+h)/ h]
=>lim h-->0 [(f(a+h)- g(a+h)) - f(a) + g(a+h)/ h]

this implies that we want to show

lim h-->0 [g(a+h) - f(a)/h = L]

I'm not sure where to go from here.


Direction 2:
<=

Assume that lim h-->0 [f(a+h) -g(a+h) / h = 0]
show lim h-->0 [f(a+h) - f(a) / h] is equal to some constant.


Take lim h-->0 [f(a+h) -g(a+h) / h = 0]:

=> lim h-->0 [f(a+h) -g(a+h) -f(a) + f(a) - g(a) + g(a) / h = 0]

=> lim h-->0 [(f(a+h)-f(a)) -(g(a+h) - g(a)) + (f(a) - g(a))/ h = 0]

but lim h-->0 [f(a+h)-f(a)/h] is just f'
and lim h-->0 [g(a+h) - g(a)/h] is -m
so, we just need to show that
f(a) = g(a) and then lim h-->0 [f(a) - g(a)/h] = 0 and f' = m
Not sure how to do that.

That's all.

Thanks for your help.
 
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  • #2
There exists g(x)=m+b(x-a) such that f and g are equal to the first order, so you need to find such g (i.e., the value of m and b) according to the diff'bility of f.
 
  • #3
for both parts?
 
  • #4
No, kekido said "such that f and g are equal to the first order" so he is talking about "if there exist g= m(x-a)+b such that f and g are equal to first order, then f is differentiable at x=a."

However, the theorem as stated is not true. For example, if f(x)= 2x if x is not 0, 1 if x= 0, and g(x)= 2x, we have f(0+h)= 2h for all non-zero h, g(0+h)= 2h for all h so
[tex]lim_{x\rightarrow a}\frac{f(0+h)-g(0+h)}{h}= 0[/itex]
but f is not differentiable at n.

You have to have some statement about what happens at x= a.
 

1. What does it mean for a function to be equal up to nth order?

When a function is said to be equal up to nth order, it means that the function and its first n derivatives are equal at a specific point. This means that the function and its derivatives have the same value, slope, concavity, and higher-order derivatives at that point.

2. How is the nth order of a function defined?

The nth order of a function is defined as the derivative of the (n-1)th order. In other words, the nth order of a function is the derivative of the (n-1)th derivative.

3. What is the importance of knowing the nth order of a function?

Knowing the nth order of a function is important in many areas of mathematics and science. It allows us to understand the behavior of a function at a specific point and make predictions about its behavior in the future. It also helps in solving differential equations and understanding the rate of change of a function.

4. What is diffbility and why is it important?

Diffbility refers to the ability of a function to have a well-defined derivative at a specific point. A function is considered diffbile if its derivative exists at that point. Knowing the diffbility of a function is important because it allows us to determine if the function is continuous and smooth at that point, and it also helps in understanding the behavior of the function.

5. Can a function be equal up to different orders at different points?

Yes, a function can be equal up to different orders at different points. This is because the function and its derivatives may have different values at different points, and therefore, the equality up to nth order may vary. It is important to specify the point at which the function is equal up to nth order when discussing this concept.

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