## 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 nth 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 Dr 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, $\huge (x+a)^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}$ 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 nth derivative of xn. Which is very true… The following simple relation holds for all n… $\huge \frac{\mathrm{d} }{\mathrm{d} x}^{n}(x^{n})=n!$ 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 Dr represents the rth derivate of xn.Hence, we can now write this as a sum, $\huge (x+a)^{n}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!}$ Or as the sum, $\huge (x+a)^{n}=\sum_{0}^{n}\frac{D_{r}x^{r}}{r!}$ So, we now have the expansion in terms of combinations as well as in terms of derivatives! $\huge (x+a)^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!}$

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#### Discuss - 2 Comments

1. rick woods says:

Great. now, can we generate the nth. der, without generating all the derivatives in betweeen

2. maria andros says:

Great work keep it coming

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