- #1
Werg22
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This question comes from how Leibniz chose his notation.
How to prove that the limit when h goes to 0 of the expression:
[tex]\frac{f(x + 2h) - f(x + h) - [f(x + h) - f(x)]}{h^{2}}[/tex]
is f''(x)?
Step 1: We know that
[tex]\frac{f(x + h) - f(x)}{h} = f'(x) + a[/tex]
Where "a" is a value that can be as small as we want, in function of h.
Step2:
Also,
It has occurred to me that first we must prove that
[tex]\frac{f(x + 2h) - f(x + h)}{h}[/tex] can be written under the form
[tex]f'(x + h) + b[/tex]
Step 3: The last condition that must be fufilled is that the limit as h goes to 0 of the expression [tex]\frac{b - a}{h}[/tex] is 0.
Step 4: That way we start with
[tex]\frac{f(x + 2h) - f(x + h) - [f(x + h) - f(x)]}{h^{2}}[/tex]
We write
[tex]\frac{f'(x + h) + b - [f'(x) + a]}{h}[/tex]
We rearange so
[tex]\frac{f'(x + h) - f'(x)}{h}+ \frac{b - a}{h}[/tex]
Now it would be clear the limit is f''(x).
The real problem is to prove step 2 and step 3... I tried but nothing occurred to me. Anyone care to try/help? Thanks in advance.
How to prove that the limit when h goes to 0 of the expression:
[tex]\frac{f(x + 2h) - f(x + h) - [f(x + h) - f(x)]}{h^{2}}[/tex]
is f''(x)?
Step 1: We know that
[tex]\frac{f(x + h) - f(x)}{h} = f'(x) + a[/tex]
Where "a" is a value that can be as small as we want, in function of h.
Step2:
Also,
It has occurred to me that first we must prove that
[tex]\frac{f(x + 2h) - f(x + h)}{h}[/tex] can be written under the form
[tex]f'(x + h) + b[/tex]
Step 3: The last condition that must be fufilled is that the limit as h goes to 0 of the expression [tex]\frac{b - a}{h}[/tex] is 0.
Step 4: That way we start with
[tex]\frac{f(x + 2h) - f(x + h) - [f(x + h) - f(x)]}{h^{2}}[/tex]
We write
[tex]\frac{f'(x + h) + b - [f'(x) + a]}{h}[/tex]
We rearange so
[tex]\frac{f'(x + h) - f'(x)}{h}+ \frac{b - a}{h}[/tex]
Now it would be clear the limit is f''(x).
The real problem is to prove step 2 and step 3... I tried but nothing occurred to me. Anyone care to try/help? Thanks in advance.
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