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Question Number 9784 by FilupSmith last updated on 04/Jan/17

For (x+y)^n :n∈Z (x+y)^n =Σ_(u=0) ^n ((n),(u) ) x^(n−u) y^n Is there a generalization for when n is non integer? e.g. (x+y)^(1/n) :n^(−1) ∉Z

$$\mathrm{For}\:\left({x}+{y}\right)^{{n}} :{n}\in\mathbb{Z} \\ $$$$\left({x}+{y}\right)^{{n}} =\underset{{u}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{u}}\end{pmatrix}\:{x}^{{n}−{u}} {y}^{{n}} \\ $$$$\: \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{generalization}\:\mathrm{for}\:\mathrm{when} \\ $$$${n}\:\mathrm{is}\:\mathrm{non}\:\mathrm{integer}? \\ $$$$\: \\ $$$$\mathrm{e}.\mathrm{g}.\:\:\:\:\left({x}+{y}\right)^{\mathrm{1}/{n}} :{n}^{−\mathrm{1}} \notin\mathbb{Z} \\ $$

Commented by FilupSmith last updated on 05/Jan/17

Taylor Series: f(x)=(x+y)^N N∈C f(x)=Σ_(u=0) ^∞ ((f^((u)) (a))/(u!))(x−a)^u a=1−y f(x)=Σ_(u=0) ^∞ ((f^((u)) (1−y))/(u!))(x+y−1)^u f(x)=(((1−y+y)^N )/(0!))(x+y+1)^0 +((N(1−y+y)^(N−1) )/(1!))(x+y−1)^1 +... ∴f(x)=1+N(x+y−1)+(1/2)N(N−1)(x+y−1)^2 +(1/4)N(N−1)(N−2)(x+y−1)^3 +... f(x)=(x+y)^N =Σ_(u=0) ^∞ (((N!∙(x+y−1)^u )/(u!∙(N−u)!))) N∉Z =Σ_(u=0) ^∞ (((Γ(N+1)∙(x+y−1)^u )/(u!∙Γ(N+1−u)))) ∴f(x)=(x+y)^N =Σ_(u=0) ^∞ (((Γ(N+1))/(u!∙Γ(N+1−u)))(x+y−1)^u ) if N∈Z =Σ_(u=0) ^∞ ( ((N),(u) )(x+y−1)^u )

$$\mathrm{Taylor}\:\mathrm{Series}: \\ $$$${f}\left({x}\right)=\left({x}+{y}\right)^{{N}} \:\:\:\:\:\:\:\:\:\:{N}\in\mathbb{C} \\ $$$${f}\left({x}\right)=\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({u}\right)} \left({a}\right)}{{u}!}\left({x}−{a}\right)^{{u}} \\ $$$$\: \\ $$$${a}=\mathrm{1}−{y} \\ $$$${f}\left({x}\right)=\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({u}\right)} \left(\mathrm{1}−{y}\right)}{{u}!}\left({x}+{y}−\mathrm{1}\right)^{{u}} \\ $$$${f}\left({x}\right)=\frac{\left(\mathrm{1}−{y}+{y}\right)^{{N}} }{\mathrm{0}!}\left({x}+{y}+\mathrm{1}\right)^{\mathrm{0}} +\frac{{N}\left(\mathrm{1}−{y}+{y}\right)^{{N}−\mathrm{1}} }{\mathrm{1}!}\left({x}+{y}−\mathrm{1}\right)^{\mathrm{1}} +... \\ $$$$\therefore{f}\left({x}\right)=\mathrm{1}+{N}\left({x}+{y}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{N}\left({N}−\mathrm{1}\right)\left({x}+{y}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{N}\left({N}−\mathrm{1}\right)\left({N}−\mathrm{2}\right)\left({x}+{y}−\mathrm{1}\right)^{\mathrm{3}} +... \\ $$$$\: \\ $$$${f}\left({x}\right)=\left({x}+{y}\right)^{{N}} =\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{N}!\centerdot\left({x}+{y}−\mathrm{1}\right)^{{u}} }{{u}!\centerdot\left({N}−{u}\right)!}\right) \\ $$$${N}\notin\mathbb{Z} \\ $$$$=\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\Gamma\left({N}+\mathrm{1}\right)\centerdot\left({x}+{y}−\mathrm{1}\right)^{{u}} }{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}\right) \\ $$$$\therefore{f}\left({x}\right)=\left({x}+{y}\right)^{{N}} =\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\Gamma\left({N}+\mathrm{1}\right)}{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}\left({x}+{y}−\mathrm{1}\right)^{{u}} \right) \\ $$$$\mathrm{if}\:{N}\in\mathbb{Z} \\ $$$$=\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\begin{pmatrix}{{N}}\\{{u}}\end{pmatrix}\left({x}+{y}−\mathrm{1}\right)^{{u}} \right) \\ $$

Commented by prakash jain last updated on 04/Jan/17

Taylor series is valid. However convergence test are still required so it will not converge for all values of x and y.

$$\mathrm{Taylor}\:\mathrm{series}\:\mathrm{is}\:\mathrm{valid}.\:\mathrm{However}\:\mathrm{convergence} \\ $$$$\mathrm{test}\:\mathrm{are}\:\mathrm{still}\:\mathrm{required}\:\mathrm{so}\:\mathrm{it}\:\mathrm{will}\:\mathrm{not} \\ $$$$\mathrm{converge}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y}. \\ $$

Commented by FilupSmith last updated on 05/Jan/17

How do you test that?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{test}\:\mathrm{that}? \\ $$

Commented by FilupSmith last updated on 05/Jan/17

f(x)=Σ_(u=0) ^∞ (((Γ(N+1))/(u!∙Γ(N+1−u)))(x+y−1)^u ) Converges if L=lim_(n→∞) ∣(a_(n+1) /a_n )∣<1 L=lim_(u→∞) ∣((((Γ(N+1))/((u+1)!∙Γ(N+1−u−1)))(x+y−1)^(u+1) )/(((Γ(N+1))/(u!∙Γ(N+1−u)))(x+y−1)^u ))∣ L=lim_(u→∞) ∣((Γ(N+1)(x+y−1)^(u+1) )/((u+1)!∙Γ(N+1−u−1)))×((u!∙Γ(N+1−u))/(Γ(N+1)(x+y−1)^u ))∣ L=lim_(u→∞) ∣(((x+y−1))/((u+1)!∙Γ(N+1−u−1)))×((u!∙Γ(N+1−u))/1)∣ L=lim_(u→∞) ∣(((x+y−1))/((u+1)∙Γ(N−u)))×((Γ(N−u+1))/1)∣ L=lim_(u→∞) ∣(((x+y−1)∙(N−u+1))/((u+1)))×(1/1)∣ L=lim_(u→∞) ∣(((x+y−1)(N−u+1))/((u+1)))∣ L′Hopital′s Law L=lim_(u→∞) ∣(((x+y−1)(−1))/1)∣ L=∣−(x+y−1)∣ If L<1, f(x) converges ∴∣x+y−1∣<1 f(x)=(x+y)^N =Σ_(u=0) ^∞ (((Γ(N+1))/(u!∙Γ(N+1−u)))(x+y−1)^u ) if: ∣x+y−1∣<1

$${f}\left({x}\right)=\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\Gamma\left({N}+\mathrm{1}\right)}{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}\left({x}+{y}−\mathrm{1}\right)^{{u}} \right) \\ $$$$\: \\ $$$$\mathrm{Converges}\:\mathrm{if}\:{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\mid<\mathrm{1} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\frac{\Gamma\left({N}+\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)!\centerdot\Gamma\left({N}+\mathrm{1}−{u}−\mathrm{1}\right)}\left({x}+{y}−\mathrm{1}\right)^{{u}+\mathrm{1}} }{\frac{\Gamma\left({N}+\mathrm{1}\right)}{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}\left({x}+{y}−\mathrm{1}\right)^{{u}} }\mid \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\Gamma\left({N}+\mathrm{1}\right)\left({x}+{y}−\mathrm{1}\right)^{{u}+\mathrm{1}} }{\left({u}+\mathrm{1}\right)!\centerdot\Gamma\left({N}+\mathrm{1}−{u}−\mathrm{1}\right)}×\frac{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}{\Gamma\left({N}+\mathrm{1}\right)\left({x}+{y}−\mathrm{1}\right)^{{u}} }\mid \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left({x}+{y}−\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)!\centerdot\Gamma\left({N}+\mathrm{1}−{u}−\mathrm{1}\right)}×\frac{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}{\mathrm{1}}\mid \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left({x}+{y}−\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)\centerdot\Gamma\left({N}−{u}\right)}×\frac{\Gamma\left({N}−{u}+\mathrm{1}\right)}{\mathrm{1}}\mid \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left({x}+{y}−\mathrm{1}\right)\centerdot\left({N}−{u}+\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{1}}\mid \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left({x}+{y}−\mathrm{1}\right)\left({N}−{u}+\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)}\mid \\ $$$${L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left({x}+{y}−\mathrm{1}\right)\left(−\mathrm{1}\right)}{\mathrm{1}}\mid \\ $$$${L}=\mid−\left({x}+{y}−\mathrm{1}\right)\mid \\ $$$$\mathrm{If}\:{L}<\mathrm{1},\:{f}\left({x}\right)\:\mathrm{converges} \\ $$$$\therefore\mid{x}+{y}−\mathrm{1}\mid<\mathrm{1} \\ $$$$\: \\ $$$${f}\left({x}\right)=\left({x}+{y}\right)^{{N}} =\underset{{u}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\Gamma\left({N}+\mathrm{1}\right)}{{u}!\centerdot\Gamma\left({N}+\mathrm{1}−{u}\right)}\left({x}+{y}−\mathrm{1}\right)^{{u}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}:\:\:\mid{x}+{y}−\mathrm{1}\mid<\mathrm{1} \\ $$

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