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$\begingroup$

$$\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}$$

$$\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}$$

$$\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}$$

Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$.

Any help will be appreciated.

  • 1
    I have noticed that α(x)+β(x)+γ(x)=e^x. and that e^(x+y)=e^x e^y. Also that α'(x)=β(x),β'(x)=γ(x) and γ'(x)=α(x). @Emre – 2017-01-16
  • 1
    you must use your sums – 2017-01-16
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    What does this mean? can you explain further? @Dr.SonnhardGraubner – 2017-01-16

2 Answers 2

1

$$\alpha(x+y) \\ =\sum_{j=0}^\infty \frac{(x+y)^{3j}}{(3j)!}\\=\sum_{j=0}^\infty\sum_{k=0}^{3j} \binom {3j} k \frac{x^ky^{3j-k}}{(3j)!}\\ =\sum_{j=0}^\infty\sum_{k=0}^{3j} \frac {(3j)!}{k!(3j-k)!} \frac{x^ky^{3j-k}}{(3j)!}\\ =\sum_{j=0}^\infty\sum_{k=0}^{3j} \frac{x^ky^{3j-k}}{k!(3j-k)!}\\ =\sum_{j,k, k\le 3j} \frac{x^{k}}{k!} \frac{y^{3j-k}}{(3j-k)!}$$

Let $i=3j-k$.
$$\sum_{j,k, k\le 3j} \frac{x^{k}}{k!} \frac{y^{3j-k}}{(3j-k)!}\\ =\sum_{3\mid i+k} \frac{x^{k}}{k!} \frac{y^{i}}{i!}$$

Let $n$ be integer part of $\frac k 3$, one of these hold: $k=3n, k=3n+2, k=3n+1$

$$\sum_{3\mid i+k} \frac{x^{k}}{k!} \frac{y^{i}}{i!}\\ =\sum_{n, 3\mid i} \frac{x^{3n}}{(3n)!} \frac{y^{i}}{i!}+ \sum_{n, 3\mid i+1} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{i}}{i!}+ \sum_{n, 3\mid i+2} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{i}}{i!}$$

Let $m$ be integer part of $\frac i 3$.

$$\sum_{n, 3\mid i} \frac{x^{3n}}{(3n)!} \frac{y^{i}}{i!}+ \sum_{n, 3\mid i+1} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{i}}{i!}+ \sum_{n, 3\mid i+2} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{i}}{i!}\\ =\sum_{n, m} \frac{x^{3n}}{(3n)!} \frac{y^{3m}}{(3m)!}+ \sum_{n, m} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{3m+1}}{(3m+1)!}+ \sum_{n, m} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{3m+2}}{(3m+2)!}\\ =\alpha(x)\alpha(y)+\beta(x)\gamma(y)+\gamma(x)\beta(y)$$.

  • 0
    is there a simpler way of showing this other than the way you have worked out? @didgogns – 2017-01-16
2

Hint: Let's assume that we have $\sigma=e^{i 2\pi /3}=-1/2+i\sqrt{3}/2$

$\sigma^3-1=0$ $$(\sigma-1)(\sigma^2+\sigma+1)=0$$

$\sigma^2+\sigma+1=0$

$$e^{ x} =\sum_{j=0}^\infty \frac{ x^{ j}}{(j)!}=1+\frac{ x^{1}}{1!}+\frac{ x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{ x^{4}}{4!}+...$$

$$e^{\sigma x} =\sum_{j=0}^\infty \frac{\sigma^{j} x^{ j}}{(j)!}=1+\frac{\sigma x^{1}}{1!}+\frac{\sigma ^2 x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{\sigma x^{4}}{4!}...$$

$$ e^{\sigma^2 x} =\sum_{j=0}^\infty \frac{\sigma^{2j} x^{ j}}{(j)!}=1+\frac{\sigma^2 x^{1}}{1!}+\frac{\sigma x^{2}}{2!}+\frac{ x^{3}}{3!}+\frac{\sigma^2 x^{4}}{4!}+...$$

$$ e^{ x} +e^{\sigma x}+e^{\sigma^2 x} =3+\frac{(1+\sigma+\sigma^2) x^{1}}{1!}+\frac{(1+\sigma+\sigma^2) x^{2}}{2!}+\frac{3 x^{3}}{3!}+\frac{(1+\sigma+\sigma^2) x^{4}}{4!}+...$$

$$ e^{ x} +e^{\sigma x}+e^{\sigma^2 x} =3(1+\frac{ x^{3}}{3!}+\frac{ x^{6}}{6!}+\frac{ x^{9}}{9!}+.....)$$

$$\alpha(x)=\frac{ e^{ x} +e^{\sigma x}+e^{\sigma^2 x}}{3} $$

And you know that

$$\beta(x)=\frac{ x^{2}}{2!}+\frac{ x^{5}}{5!}+\frac{ x^{8}}{8!}+.....=\alpha'(x)=\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3}$$

$$\beta(x)=\alpha'(x)=\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3}$$

$$\gamma(x)=\frac{ x^{1}}{1!}+\frac{ x^{4}}{4!}+\frac{ x^{7}}{7!}+.....=\alpha''(x)=\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3}$$

$$\gamma(x)=\alpha''(x)=\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3}$$

Now you can use them prove your identity

$$\alpha(x)\alpha(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)=$$ $$(\frac{ e^{ x} +e^{\sigma x}+e^{\sigma^2 x}}{3})(\frac{ e^{ y} +e^{\sigma y}+e^{\sigma^2 y}}{3})+(\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3})(\frac{ e^{ y} +\sigma^2 e^{\sigma y}+\sigma e^{\sigma^2 y}}{3})+(\frac{ e^{ y} +\sigma e^{\sigma y}+\sigma^2 e^{\sigma^2 y}}{3})(\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3}) $$

Can you continue after that??

  • 0
    how do I use this to prove my identity? – 2017-01-16