The Derivative & The Binomial Theorem

If we observe closely, we find that the various branches of mathematics are all linked together in some way or the other. I show here one such observation. The binomial theorem can actually be expressed in terms of the derivatives of xⁿ instead of the use of combinations. Lets start with the standard representation of the binomial theorm, $large (x+a)^{n}=x^{n}+ ^{n}C_{1}ax^{n-1}+ ^{n}C_{2}a^{2}x^{n-2}+^{n}C_{3}a^{3}x^{n-3}+...^{n}C_{n}a^{n}$ We could then rewrite this as a sum, $huge (x+a)^{n}=sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}$ Another way of writing the same thing would be, $large (x+a)^{n}=x^{n}+nax^{n-1}+frac{n(n-1)}{1.2}a^{2}x^{n-2}+frac{n(n-1)(n-2)}{1.2.3}a^{3}x^{n-3}+...a^{n}$ We observe here that the equation can be rewritten in terms of the derivatives of xⁿ. The coefficient of a in the second terms is the first derivative of xⁿ, similarily the coefficient of a²/2! in the second term is the second derivative of x… Lets look at the last term of the expansion, The coefficient of xⁿ/n! should now be the n^th derivative of xⁿ. Which is very true… The following simple relation holds for all n… $large frac{mathrm{d} }{mathrm{d} x}^{n}(x^{n})=n!$ Hence, the binomial expansion can now be written in terms of derivatives! We have, $large (x+a)^{n}=x^{n}+aD_{1}+frac{D_{2}}{2!}a^{2}+frac{D_{3}}{3!}a^{3}...frac{D_{n}}{n!}a^{n}$ where D_r represents the rth derivate of xⁿ. Hence, we can now write this as a sum, http://latex.codecogs.com/gif.latex?huge%20(x+a)^{n}=sum_{0}^{n}frac{D_{r}a^{r}}{r!} So, we now have the expansion in terms of combinations as well as in terms of derivatives! $huge (x+a)^{n}=sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}=sum_{0}^{n}frac{D_{r}a^{r}}{r!}$ ⁿ instead of the use of combinations. Lets start with the standard representation of the binomial theorm, $(x+a)^{n}=x^{n}+ ^{n}C_{1}ax^{n-1}+ ^{n}C_{2}a^{2}x^{n-2}+^{n}C_{3}a^{3}x^{n-3}+...^{n}C_{n}a^{n}$ We could then rewrite this as a sum, $(x+a)^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}$ Another way of writing the same thing would be, $(x+a)^{n}=x^{n}+nax^{n-1}+\frac{n(n-1)}{1.2}a^{2}x^{n-2}+\frac{n(n-1)(n-2)}{1.2.3}a^{3}x^{n-3}+...a^{n}$ We observe here that the equation can be rewritten in terms of the derivatives of xⁿ. The coefficient of a in the second terms is the first derivative of xⁿ, similarily the coefficient of a²/2! in the second term is the second derivative of x… Lets look at the last term of the expansion, The coefficient of xⁿ/n! should now be the n^th derivative of xⁿ. Which is very true… The following simple relation holds for all n… $\frac{\mathrm{d} }{\mathrm{d} x}^{n}(x^{n})=n!$ Hence, the binomial expansion can now be written in terms of derivatives! We have, $(x+a)^{n}=x^{n}+aD_{1}+\frac{D_{2}}{2!}a^{2}+\frac{D_{3}}{3!}a^{3}...\frac{D_{n}}{n!}a^{n}$ where D_r represents the rth derivate of xⁿ.Hence, we can now write this as a sum, $(x+a)^{n}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!}$ Or as the sum, $(x+a)^{n}=\sum_{0}^{n}\frac{D_{r}x^{r}}{r!}$ So, we now have the expansion in terms of combinations as well as in terms of derivatives! $(x+a)^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!}$