User:Ed g2s/Maths

I have put this 'proof' here as an interesting result I stumbled upon a while back. I have never seen it in publication (but then again I don't read too many books), nor have I had an expert check it. The algebra is watertight, but some of my assumptions may require more proof. Nonetheless it is an interesting take on a subject you probably don't give much thought too. (I recommend reading Order 2/Method 2 and Order N) Ed g2s 02:28, 5 Jul 2004 (UTC)

Some functions can be differentiated without the need for Newton's difference quotient and without having to study limits.

By constructing suitable tangent lines, all polynomials can be differentiated.

Order 1

$\ f(x)=ax+b$

The gradient of this line is $\ a$ which is therefore the derivative at any point.

$\ f^{\prime }(x)=a$

Order 2

Method 1

$\ f(x)=ax^{2}+bx+c$

Consider a line $\ g(x)$ such that it intersects the quadratic twice at $\ x=x_{1}$ (a double root). This is the tangent to the curve at this point.

Subtracting the two equations will give a quadratic with roots $\ x_{1},x_{1}$, meaning it will be of the form:

$\ f(x)-g(x)=A(x-x_{1})(x-x_{1})$

where A is a non-zero constant. We can choose $\ A=a$.

$\ g(x)=f(x)-a(x-x_{1})^{2}$

$\Rightarrow g(x)=ax^{2}+bx+c-ax^{2}+2ax_{1}x-ax_{1}^{2}$

$\Rightarrow g(x)=(2ax_{1}+b)x+c-ax_{1}^{2}$

The gradient of this line is $\ 2ax_{1}+b$. This is also the rate of change of the function $\ f(x)$ at the point $\ x=x_{1}$.

$\ f^{\prime }(x_{1})=g^{\prime }(x_{1})=2ax_{1}+b$

To conclude:

$\ f(x)=ax^{2}+bx+c$ $\Rightarrow f^{\prime }(x)=2ax+b$

Method 2

$\ f(x)=ax^{2}+bx+c$

Let us imagine a tangent line that intersects the above quadratic once and only once where $\ x=x_{1}$. This tangent line will be of the form:

$\ g(x)=mx+n$

The difference of the two functions:

$\ f(x)-g(x)=ax^{2}+(b-m)x+(c-n)$

will be a quadratic that has a double root at

$\ x=x_{1}$

$\ f(x_{1})-g(x_{1})=ax_{1}^{2}+(b-m)x_{1}+(c-n)=0$

$\Rightarrow c=n-(b-m)x_{1}-ax_{1}^{2}\ \ \ (*)$

using the quadratic equation:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}=x_{1}$

As $\ x_{1}$ is a double root, the discriminant of the equation is equal to zero.

$\ b^{2}-4ac=0$ $\Rightarrow (b-m)^{2}-4a(c-n)=0$

Substituting in $\ c$ from $\ (*)$:

$m^{2}-2bm+b^{2}-4a(n-(b-m)x_{1}-ax_{1}^{2}-n)=0$ $\Rightarrow m^{2}-(4ax_{1}+2b)m+(b^{2}-4a^{2}x_{1}^{2})$

Solving this as a quadratic equation in $\ m$:

$m={\frac {(4ax_{1}+2b)\pm {\sqrt {(4ax_{1}+2b)^{2}-4(b^{2}-4a^{2}x_{1}^{2})}}}{2}}$ $\ m=2ax_{1}+b$

Which gradient of the quadratic at $\ x=x_{1}$. To conclude:

$\ f(x)=ax^{2}+bx+c$ $\Rightarrow f^{\prime }(x)=2ax+b$

Order 3

$\ f(x)=ax^{3}+bx^{2}+cx+d$

Consider a quadratic $\ g(x)$ such that it intersects the cubic twice at $\ x=x_{1}$ (a double root) and once at $\ x=x_{2}$. You can think of this quadratic as a tangent curve.

Subtracting the two equations will give a cubic with roots $\ x_{1},x_{1},x_{2}$, meaning it will be of the form:

$\ f(x)-g(x)=A(x-x_{1})(x-x_{1})(x-x_{2})$

where A is a non-zero constant. We can choose $\ A=a$.

$\ g(x)=f(x)-a(x-x_{1})^{2}(x-x_{2})$ $\Rightarrow g(x)=(a(2x_{1}+x_{2})+b)x^{2}+(c-ax_{1}(x_{1}+2x_{2}))x+ax_{1}^{2}x_{2}+d$

Using the result for Order 2 polynomials:

$\ g^{\prime }(x)=2(a(2x_{1}+x_{2})+b)x-ax_{1}(x_{1}+2x_{2})+c$

As $\ f(x)$ and $\ g(x)$ share a tangent at $\ x=x_{1}$

$\ f^{\prime }(x_{1})=g^{\prime }(x_{1})=2(a(2x_{1}+x_{2})+b)x_{1}-ax_{1}(x_{1}+2x_{2})+c$ $\ f^{\prime }(x_{1})=3ax_{1}^{2}+2bx_{1}+c$

To conclude:

$\ f(x)=ax^{3}+bx^{2}+cx+d$ $\Rightarrow f^{\prime }(x)=3ax^{2}+2bx+c$

Order N

To prove differentiation for a general polynomial an inductive proof can be used.

Assume true for $\ n=k$:

${\frac {d}{dx}}\sum _{r=0}^{k}a_{r}x^{r}=\sum _{r=0}^{k}ra_{r}x^{r-1}$

If we can show it is true for $\ n=k+1$ using the above assumption then we will have shown it true for all $\ n\in \mathbb {N}$ as we've shown it true for the first few cases.

We need to show that:

$\ f(x)=Ax^{k+1}+\sum _{r=0}^{k}a_{r}x^{r}$

differentiates to:

$\ f^{\prime }(x)=A(k+1)x^{k}+\sum _{r=0}^{k}ra_{r}x^{r-1}$

Consider a polynomial $\ g(x)$ order $\ k$ such that it intersects $\ f(x)$ twice at $\ x=x_{1}$ (a double root) and for simplicity the remaining $\ k-1$ intersections are all at $\ x=x_{2}$.

Subtracting the two equations will give the following result.

$\ f(x)-g(x)=A(x-x_{1})^{2}(x-x_{2})^{k-1}$ $\Rightarrow g(x)=Ax^{k+1}-A(x-x_{1})^{2}(x-x_{2})^{k-1}+\sum _{r=0}^{k}a_{r}x^{r}$

Using the binomial theorem:

$g(x)=Ax^{k+1}-A(x^{2}-2x_{1}x+x_{1}^{2})\sum _{r=0}^{k-1}(^{k-1}C_{r}(-x_{2})^{k-1-r}x^{r})+\sum _{r=0}^{k}a_{r}x^{r}$

Taking out the last term of the sum (where $\ r=k-1$)

$g(x)=Ax^{k+1}-A(x^{2}-2x_{1}x+x_{1}^{2})x^{k-1}-\sum _{r=0}^{k-2}\ \left(\ ^{k-1}C_{r}(-x_{2})^{k-1-r}x^{r}\cdot A(x^{2}-2x_{1}x+x_{1}^{2})\right)+\sum _{r=0}^{k}a_{r}x^{r}$ $\Rightarrow g(x)=2Ax_{1}x^{k}-Ax_{1}^{2}x^{k-1}-A\sum _{r=0}^{k-2}\ \left(\ ^{k-1}C_{r}(-x_{2})^{k-1-r}(x^{r+2}-2x_{1}x^{r+1}+x_{1}^{2}x^{r})\right)+\sum _{r=0}^{k}a_{r}x^{r}$

As all the terms are now of order $\ k$ or less we can differentiate the expression using the assumption above:

$g^{\prime }(x)=2Akx_{1}x^{k-1}-A(k-1)x_{1}^{2}x^{k-2}-A\sum _{r=0}^{k-2}\ \left(\ ^{k-1}C_{r}(-x_{2})^{k-1-r}\left((r+2)x^{r+1}-2(r+1)x_{1}x^{r}+rx_{1}^{2}x^{r-1}\right)\ \right)+\sum _{r=0}^{k}ra_{r}x^{r-1}$

As $\ f(x)$ and $\ g(x)$ share a tangent at $\ x=x_{1}$

$f^{\prime }(x_{1})=g^{\prime }(x_{1})=2Akx^{k}-Akx^{k}+Ax^{k}-A\sum _{r=0}^{k-2}\ \left(\ ^{k-1}C_{r}(-x_{2})^{k-1-r}(r+2-2(r+1)+r)x^{r+2}\right)+\sum _{r=0}^{k}ra_{r}x^{r-1}$

As $\ (r+2-2(r+1)+r)=0$ the entire sum $\ A\sum _{r=0}^{k-2}\ \left(...\right)$ cancels out giving:

$\ f^{\prime }(x)=A(k+1)x^{k}+\sum _{r=0}^{k}ra_{r}x^{r-1}$

By induction we have proved the result

${\frac {d}{dx}}\sum _{r=0}^{n}a_{r}x^{r}=\sum _{r=0}^{n}ra_{r}x^{r-1}$

is true for all $\ n\in \mathbb {N}$.