# Higher Order derivative representation

1. Oct 12, 2011

### pari777

Hi everybody, I have a question:
We know that the geometrical representation of 1st order derivative is the slope of a function. Then what is the geometrical representation of derivatives having order more than 1? I mean what does it actually represent in a function? Please some body clear my doubt.

2. Oct 12, 2011

### Eynstone

The geometric interpretation of the second derivative can be convexity at that point, but I have no clue about higher derivatives.

3. Oct 15, 2011

### pari777

ummmm...what's convexity at a point?

4. Oct 15, 2011

### hamood_rak

Convexity of a shape is that it should not have a "dent" , e.g., a circle or regular polygons are convex figures. The general idea is that for any two points that lie in the set (or "shape") the line segment that joins the two points should also be in the set.

For functions this means that if a function is "concave up" or "convex" then the the line segment joining any two points $(a,f(a))$ and $(b, f(x))$, on the graph of $f(x)$ must lie at or above the points of the graph that are between $(a,f(a))$ and $(b, f(b))$. Roughly there should be a "trough" between the two points on the graph. For concave function the opposite is true, with the "trough" replaced by crest.

Mathematically the condition is given by:

$f(x_1 t + (1-t)x_2) \leq f(x_1) t +(1-t)f(x_2) \, for \, x_1 \, and \, x_2 \in Domain(f) \, and \, t \in [0, 1]$

Last edited: Oct 15, 2011
5. Oct 18, 2011

### pari777

ok, thanks to both of you.
but tell me , why it is only convexity and not concavity for 2nd order derivative.

6. Oct 18, 2011

### HallsofIvy

Either one. The "concavity" of a curve is just the negative of the "convexity".

7. Oct 19, 2011

### pari777

ok. thanks.
But what about the other higher derivatives?

8. Oct 19, 2011

### JJacquelin

Consider the graphical representation of a function f(x) around the point (x0, y0) where y0=f(x0)
Series expansion of f(x) around (x0,y0) is :
y(x) = y0 + (x-x0)f'(x0) + (1/2)(x-x0)²f''(x0) + (1/6)(x-x0)^3 f'''(x0) +...

The first derivative of f(x) allows to draw the straight line tangent to the curve, which equation is : y(x) = y0 + (x-x0)f'(x0)
y = a x +b where the slope is a = f'(x0) and b = y0 -x0 f'(x0)

The second derivative of f(x) allows to draw a circle tangent to the curve, called osculating circle, which equation is : y(x) = y0 + (x-x0)f'(x0) + (1/2)(x-x0)²f''(x0)
The radius is related to what is called "curvature" (with sign + or- depending on the position of the circle relatively to the curve)

The third derivative allows to draw a cubic curve tangent to the curve f(x). The equation of the cubic curve is : y(x) = y0 + (x-x0)f'(x0) + (1/2)(x-x0)²f''(x0) + (1/6)(x-x0)^3 f'''(x0)

The next derivative allows to draw a cartic curve tangent to f(x). The equation of the quartic includes the next term of the series : +(1/4!)(x-x0)^4f''''(x0)

And so on... That way, the successive derivatives and the related tangent curves (straight line, then circle, then cubic curve, then quartic curve, etc...) are closer and closer to the f(x) curve.

9. Oct 19, 2011

### pari777

the above series is Taylor series, isn't it?
you mean derivatives depend on this series? Like when the order is 2, we have 2 terms from taylor's series. When it is of order 3, we have 3 terms of taylor series and likewise?

10. Oct 19, 2011

### pari777

will these successive tangent curves touch the function f(x) curve at some nth order of the derivative or will they tend to touch but will actually never touch the f(x) curve?