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Solve $(1 + x)y' = py;\ \ \ y(0) = 1$, where $p$ is an arbitrary constant.

First I plugged in the guess $y = \sum_{n = 0}^\infty a_n x^n$:

$(1 + x)(\sum_{n = 0}^\infty a_n x^n)' = p\sum_{n = 0}^\infty a_n x^n$

Then I expanded the derivative and multiplication:

$\sum_{n = 0}^\infty n a_n x^{n - 1} + \sum_{n = 0}^\infty n a_n x^n = p\sum_{n = 0}^\infty a_n x^n$

Then I shifted the left index (the first term yielding $0$ allows the lower bound to remain $0$) and algebraically combined the summations:

$\sum_{n = 0}^\infty (n + 1)a_{n + 1} x^n + (n - p)a_n x^n = 0$

This leads to the following recurrence relation:

$a_{n + 1} = \frac{p - n}{n + 1}a_n$

Thus for various values of $n$:

$a_1 = p a_0$, $a_2 = \frac{p(p - 1)}{2}a_0$, $a_3 = \frac{p(p - 1)(p - 2)}{6} a_0$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$y = \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!} = \sum_{n = 0}^\infty a_0 e^x p^{\underline n}$

Solving the initial value problem:

$1 = \sum_{n = 0}^\infty a_0 e^0 p^{\underline n} \implies a_0 = \frac{1}{\sum_{n = 0}^\infty p^{\underline n}}$

My final solution is:

$y = \frac{\sum_{n = 0}^\infty e^x p^{\underline n}}{\sum_{n = 0}^\infty p^{\underline n}}$

However, the answer is supposed to be $y = (1 + x)^p$. Are these identical, or did I make an error somewhere?

user10478
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2 Answers2

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Your solution $$ y = \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!} $$ is mostly the binomial series for $a_0\,(1+x)^p$, assuming you know what you are doing (it looks like you are treatng $p$ as an integer, which it isn't necessarily according to the question as you posted it).

The "equality" $$ \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!}= \sum_{n = 0}^\infty a_0 e^x p^{\underline n} $$ is not true; it's missing the $n!$ in the denominator on the right.

Martin Argerami
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  • I was using the falling factorial $p^{\underline n}$, not the regular exponential $p^n$, in the RHS of that equality. – user10478 Jan 17 '21 at 02:55
  • I see what you mean. But you wrote $$\frac{p!}{n!(p-n)!},$$ which makes no sense when $p$ is not a positive integer. It is fine to use the falling factorial, but you should have written instead ${p\choose n}$, which is the standard notation. – Martin Argerami Jan 17 '21 at 04:24
  • Right, so the intermediate steps with the factorials aren't as general as they should be, although ultimately I believe the falling factorial operator can be extended to non-integers the same way the choose operator can. I chose falling factorial instead of choose because it allows $x^n$ to be simplified via reverse Maclaurin expansion to $e^x$. – user10478 Jan 17 '21 at 05:39
  • No, that's not it. Your deduction is fine. You correctly found that $$a_n=\frac{p^{\underline n}}{n!}.$$ But then when you wrote the solution you changed it first to $$\frac{p!}{n!(p-n)!}$$ and then to $p^{\underline n}$. – Martin Argerami Jan 17 '21 at 11:54
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Hint: You know that the solution (by separation of variables) is:

$y(x)=(x+1)^p$

Now try to use the binomial theorem, expand the above, and equate coefficients. You will see that:

$a_0=1$

$a_1=pa_0=p= \binom{p}{1}$

$a_2=\frac{p(p-1)}{2}a_0= \binom{p}{2}$

. . .

$a_k=\binom{p}{k}$

As the others have pointed out, $a_k=0$ for $k\geq p+1$. This is because the solution is a polynomial.

So your recursion is correct, but after you applied the exponential Taylor series, things took a different turn.

Moral of the story: If you have an easier method, use it to double check your work :)

Edit: The above assumes that $p$ is an integer. If $p$ is an arbitrary scalar value (i.e. real, or even complex) it turns out we can in fact use a similar formula, we just have to be careful of how we are computing the binomial coefficients.

Indeed, we have that for arbitrary $p$,

$$y(x)= \sum_{k=0}^{\infty}\binom{p}{k}x^k$$ for $|x|<1$ and $x,p \in \mathbb{R}$

I did a little investigating, and found this to be helpful: Generalized Binomial Expansion of $\left(1+x \right )^{y}$

Edit: Looks like there are multiple formulas for the "generalized binomial coefficients", but digging into Wolfram Mathworld, I came across:

$$\binom{p}{k}:= \frac{(p)_k}{k!}=\frac{p(p-1)(p-2)\cdots(p-k+1)}{k(k-1)\cdots 2 \cdot 1}$$

Look at formulas $(3),(4)$ in:

https://mathworld.wolfram.com/BinomialCoefficient.html

Vegeta the Prince of Saiyans
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  • Hmm, your solution $y(x)= \sum_{k=0}^{\infty}\binom{p}{k}x^k$ does not seem to satisfy the initial condition. By applying formula $(3)$ from your link to your solution and then noticing the Maclaurin expansion for $e^x$, it looks like your solution becomes $\sum_{n = 0}^\infty e^x p^{\underline n}$, which would be my solution except that yours has $a_0 = 1$ whereas mine has $a_0 = \frac{1}{\sum_{n = 0}^\infty p^{\underline n}}$, right? – user10478 Jan 17 '21 at 05:27
  • But the original assumption was that $y(x)= \sum_{k=0}^{\infty}a_{k}x^k$, correct? So if $y(0)=1$, then substituting in $y(x)= \sum_{k=0}^{\infty}a_{k}x^k:=a_0+a_1x+a_2x^2+\cdots$, we get $y(0)=a_0=1$. That's why I had $a_0=1$. – Vegeta the Prince of Saiyans Jan 17 '21 at 05:51
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    I sort of see the logic behind that, but doesn't plugging the initial condition into your solution break it? It would be saying $1 = \sum_{k=0}^{\infty}\binom{p}{k}0^k$, which is clearly not true. – user10478 Jan 17 '21 at 06:41
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    It looks like we both have a point, as there really isn't a best way to define $0^0$. Look at this post, for example: https://math.stackexchange.com/questions/1323868/can-forall-x-xe0-s0-be-one-of-axioms-for-a-theory-of-arithmetic?noredirect=1&lq=1 – Vegeta the Prince of Saiyans Jan 17 '21 at 08:07
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    or this one: https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Vegeta the Prince of Saiyans Jan 17 '21 at 08:07
  • That's an interesting observation. Suppose we look at the special case where $p$ is an arbitrary natural number (even though it is actually an arbitrary real or perhaps complex number). Here, $\sum_{k = 0}^\infty {p \choose k} x^k$ becomes $\sum_{k = 0}^\infty \frac{p!}{(p - k)!} \frac{x^k}{k!}$ which becomes $\sum_{k = 0}^\infty \frac{p!}{(p - k)!} e^x$ (the remaining factor is just the falling factorial but I'll leave it expanded). In this form, it is unambiguous that $y(0) = \sum_{k = 0}^\infty \frac{p!}{(p - k)!}$, which does not meet the condition of being identical to $1$ for all $p$. – user10478 Jan 17 '21 at 17:15