Question

Use the following steps to prove \(18\)

a. Let \(g(x) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n}} \). Differentiate this series to show that \(g'(x) = \frac{{kg(x)}}{{1 + x}}\;\;\; - 1 < x < 1\)

b. Let \(h(x) = {(1 + x)^{ - k}}g(x)\) and show that \(h'(x) = 0\).

c. Deduce that \(g(x) = {(1 + x)^k}\).

Short answer

a. Let \(g(x) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n}} \). Differentiate this series to show that \(g'(x) = \frac{{kg(x)}}{{1 + x}}\;\;\; - 1 < x < 1\)

Write the equation to prove as \((1 + x)g'(x) = kg(x)\), multiply \((1 + x)g'(x)\). Then do some rearranging and factoring to turn it into the form of \(kg(x)\)

Step-by-step solution

Step 1: Table 1 and convergence

From table 1

\(g(x) = \sum\limits_{n = 0}^\infty {\left({\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + \frac{{k(k - 1)(k - 2)(k -3)}}{{4!}}{x^4}.....} \)

The series converges for \( - 1 < x < 1\). If x is not in this range then some of the manipulations to be done might not be valid.

Everything that comes from \(\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right)\) is a constant.

\(g'(x) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right)n{x^{n - 1}} = 0 + k + \frac{{2k(k - 1)}}{{2!}}x + \frac{{3k(k - 1)(k - 2)}}{{3!}}{x^2} + \frac{{4k(k - 1)(k - 2)(k - 3)}}{{4!}}{x^3} + .....} \)\(g'(x) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right)n{x^{n - 1}} = k + k(k - 1)x + \frac{{k(k - 1)(k - 2)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)(k - 3)}}{{3!}}{x^3} + .....} \)

Step 2: Further assumption and simplification

\(\begin{array}{l}If\;g'(x) = \frac{{kg(x)}}{{1 + x}}\\then\;(1 + x)g'(x) = kx\end{array}\)\(\begin{array}{l}(1 + x)g'(x) = 0 + k + \frac{{2k(k - 1)}}{{2!}}x + \frac{{3k(k - 1)(k - 2)}}{{3!}}{x^2} + \frac{{4k(k - 1)(k - 2)(k - 3)}}{{4!}}{x^3} + ..... + kx\\ + k(k - 1){x^2} + \frac{{k(k - 1)(k - 2)}}{{2!}}{x^3} + ....\end{array}\)

Rearrange into ascending order of exponents

\(\begin{array}{l}(1 + x)g'(x) = k + kx + k(k - 1)x + k(k - 1){x^2} + \frac{{k(k - 1)(k - 2)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{2!}}{x^3}\\ + \frac{{k(k - 1)(k - 2)(k - 3)}}{{3!}}{x^3} + .....\end{array}\)

Make common denominators to combine terms with the same degree

\(\begin{array}{l}(1 + x)g'(x) = k + kx + k(k - 1)x + \frac{{2k(k - 1){x^2} + k(k - 1)(k - 2)}}{{2!}}{x^2}\\ + \frac{{3k(k - 1)(k - 2) + k(k - 1)(k - 2)(k - 3)}}{{3!}}{x^3} + .........\end{array}\)

Step 3: Simplify the above equation

From terms \(2\;\& \;3\) factor out \(kx\). For the fraction with \(2!\), factor out \(k(k - 1){x^2}\) from the two terms in the numerator. For the fraction with \(3!\), factor out \(k(k - 1)(k - 2){x^3}\) etc.

\((1 + x)g'(x) = k + kx[1 + (k - 1)] + \frac{{k(k - 1){x^2}[2 + (k - 2)]}}{{2!}} + \frac{{k(k - 1)(k - 2){x^3}[3 + (k - 3)]}}{{3!}} + .........\)\(\begin{array}{l}(1 + x)g'(x) = k + kx[k] + \frac{{k(k - 1){x^2}[k]}}{{2!}} + \frac{{k(k - 1)(k - 2){x^3}[k]}}{{3!}} + .........\\(1 + x)g'(x) = k\left[ {1 + kx + \frac{{k(k - 1){x^2}}}{{2!}} + \frac{{k(k - 1)(k - 2){x^3}}}{{3!}} + .........} \right]\\(1 + x)g'(x) = kg(x)\end{array}\)

b.  Let \(h(x) = {(1 + x)^{ - k}}g(x)\) and show that \(h'(x) = 0\).Hence proved that if \(h(x) = {(1 + x)^{ - k}}g(x)\) then \(h'(x) = 0\)Step 1: Product rule of derivative

First we need to use the product rule of derivative

\(\begin{array}{l}h'(x) = \frac{d}{{dx}}[{(1 + x)^{ - k}}g(x)]\\h'(x) = \frac{d}{{dx}}[{(1 + x)^{ - k}}]g(x) + {(1 + x)^{ - k}}\frac{d}{{dx}}[g(x)]\end{array}\)

Step 2: Further  simplification

First we made use of the product rule of derivative and now we will use the expression \(g'(x)\) obtained in the part (a) to get the answer

\(\begin{array}{l} = ( - k){(1 + x)^{ - k - 1}}g(x) + {(1 + x)^{ - k}}g'(x)\\ = - \frac{{kg(x)}}{{{{(1 + x)}^{k + 1}}}} + \frac{1}{{{{(1 + x)}^k}}}\left( {\frac{{kg(x)}}{{1 + x}}} \right)\\ = - \frac{{kg(x)}}{{{{(1 + x)}^{k + 1}}}} + \frac{{kg(x)}}{{{{(1 + x)}^{k + 1}}}}\\ = 0\end{array}\)

Hence proved that if \(h(x) = {(1 + x)^{ - k}}g(x)\)then \(h'(x) = 0\).

 c. Deduce that \(g(x) = {(1 + x)^k}\).Hence proved that \(g(x) = {(1 + x)^k}\) Step 1: Use equation from part b

Since \(h'(x) = 0\)for all values of x, we can say that it is a constant function

\(h'(x) = C\)

Therefore by using the equation in part b of this solution we can write

\(C = {(1 + x)^{ - k}}g(x)\)

Step 2: Further  simplification for g(x)

\(g(x) = C{(1 + x)^k} - - - - (eqn1)\)

It is given that

\(g(x) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n}} \).

Substituting x=0 in the above equation will give us

\(g(0) = 1 + \sum\limits_{n = 1}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){0^n} = 1} \)

Substitute this in equation 1

\(\begin{array}{l}g(0) = C{(1 + 0)^k}\\1 = C\end{array}\)

Substitute back the value of C to get

\(g(x) = {(1 + x)^k}\)