# How can I explain this?

• #1
155
7
I am tutoring a kid who just started on calculus. Today he asked me if there is a way to expand this:
(a+b)^0.5

I said as far as I know, there is no way, but then the kid said:

f'(x^0.5)= ( (x+h)^0.5-x^0.5 ) / h = 0.5 x ^ -0.5;
(x + h)^0.5 = h(0.5 x ^ -0.5) + x^0.5;

and then, he proposed that x and h can be replaced by other variables, such as a and b, and the reason why the equation (a + b)^0.5 = b(0.5 a ^ -0.5) + a^0.5 is false is because h in the original equation approximated zero, and any term with h in it at the original derivative is removed, therefore, the "actual" way (a+b)^0.5 should be expanded is:

b(0.5 a ^ -0.5) + a^0.5 + c; where c is a series of functions in terms of a and b

I suspect that some of his logic is fallacious when doing the algebra to the derivative, because the equality is only true with h-->0, then again I am not very far into the logic of maths myself, so is there a way to demonstrate that this is wrong? I know the general formula for binomial expansion with integer exponents, but I don't have the proof, so maybe that will help if you can give it to me too.

• #2
andrewkirk
Homework Helper
Gold Member
3,900
1,464
There are two errors in the first line. The first one, which you identified, is that the limit as ##h\to 0## has not been taken, and the equation does not hold if the limit is not taken.

Secondly ##\frac{(x+h)^{0.5}-x^{0.5}}{h}\neq 0.5x^{0.5}##.

I suggest you ask him to justify his reasoning to you by working through the steps above one at a time. Working through and finding his own errors will be a more useful experience than having somebody else point them out for him.

One could use a Taylor series to expand the formula, but if he's just starting calculus then Taylor series may be a bit too advanced.

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