This question comes from how Leibniz chose his notation.(adsbygoogle = window.adsbygoogle || []).push({});

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 occured 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 occured to me. Anyone care to try/help? Thanks in advance.

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# Leibniz notation, need clarification

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