# Numerical estimate of slope for y=x^2 works with any delta x?

I've just discovered that I can use Euler's method to estimating the slope of $$y=(x + c)^2$$ with perfect accuracy, regardless of what $$\Delta x$$ is used. What causes this?

Does it happen with other functions, what must they look like for it to happen?

I am using this method to estimate the slope, $$m$$ $$m \approx \frac{(x+\Delta x +c)^2-(x-\Delta x + c)^2}{2\Delta x}$$

Doesn't matter what $$\Delta x$$ is, the slope is always correct ...I would expect this only to work for tiny $$\Delta x$$, but works with large ones too.

• Just expand everything and after simple computations you'll receive exact answer since $\Delta x$ cancels. Apr 20, 2020 at 1:37
• @pw1822 If you aren't already in the process of doing so, you should write & post an answer based on your comment. Apr 20, 2020 at 1:39

In the quadratic function $$y =x^2$$, you can very easily show that the secant line between points $$(a, a^2)$$ and $$(b, b^2)$$ where $$b\neq a$$ is parallel to the tangent at the midpoint $$(\frac 12(a+b), \frac 14(a+b)^2)$$.
The slope of the secant is $$\frac{b^2 - a^2}{b-a} =b+a$$. Slope of the tangent at the midpoint (by calculus) is $$(2)(\frac 12(a+b)) = a+b$$, so they're equal and the secant is parallel to the tangent.
That's the reason for your observation. Adding a constant, or using a constant multiplier, etc. won't change the result so it applies for any function of the form $$y = k(x+c)^2 + m$$.
Let's assume $$c=0$$ (other cases are just a horizontal translation of this case). Then $$\frac{(x+\Delta x)^2-(x-\Delta x)^2}{2\Delta x}=\frac{4x\Delta x}{2\Delta x}=2x.$$ This is exactly the derivative of $$x^2$$, so this approximation is exact. This does not, of course, work for all functions $$f(x)$$. If in fact $$\frac{1}{2\Delta x}\left(f(x+\Delta x)-f(x-\Delta x)\right)=f'(x)$$ for all $$\Delta x$$, then choosing $$\Delta x=x$$ we see $$\frac{1}{2x}\left(f(2x)-f(0)\right)=f'(x)$$ for all $$x$$. You can try to solve this differential equation to find all functions $$f$$ for which the condition is true.
Edit: assuming analyticity, here's how you can do so. Express $$f(x)$$ as the power series $$\sum_{n\geq0}a_nx^n$$. The left hand side of the equation is the same as $$\frac{1}{2x}(f(2x)-f(0))=\sum_{n\geq0}2^na_{n+1}x^n$$ whereas the right hand side is $$f'(x)=\sum_{n\geq0}(n+1)a_{n+1}x^n.$$ Matching coefficients, we must have for each $$n$$ that $$2^na_{n+1}=(n+1)a_{n+1}$$, so either $$a_{n+1}=0$$ or $$n+1=2^n$$, in which case $$n=0,1$$. We therefore deduce that only coefficients that can be nonzero are $$a_0,a_1$$ and $$a_2$$, and so $$f(x)$$ is quadratic.