5
$\begingroup$

I am reading slides from my lecture and saw this equality: $$ \frac{1}{2}(e^{3x} - e^{2x} - e^x + \color{red}{1}) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} $$

but in my opinion it should be $$ \frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \color{red}{\frac{1}{2}} $$ where $ \color{red}{1}$ has been lost?

$\endgroup$
  • 3
    $\begingroup$ Are $x$ and $r$ the same variable? $\endgroup$ – José Carlos Santos Aug 11 '19 at 11:23
4
$\begingroup$

You are correct, it should be $$\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \frac{1}{2}.$$ Note that, since the left-hand side is zero for $x=0$, you may also write $$\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge \color{red}{1}}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!}$$ where starting index in the sum is now $1$.

$\endgroup$
3
$\begingroup$

You are correct the $1$ term was "lost". For each $e^{nx}$ term on the left, there's a corresponding $\sum_{r\ge 0}n^r\left(\frac{x^r}{r!}\right)$ (I assume your "$t$" is meant to be "$x$") term on the right. Thus, the $\frac{1}{2}(e^3x - e^{2x} - e^x)$ is accommodated by the terms on the right, so the $ + 1$ term is, as you state, not included and, thus, should give an extra $\frac{1}{2}$ term.

An easy way to check is to use $x = 0$ on the left. This gives $0$ but, on the right with $x^0 = 1$, you get $\frac{1}{2}(1 - 1 - 1) = -\frac{1}{2}$ instead.

$\endgroup$
2
$\begingroup$

The value of the function at $0$ is $0$, so the constant term of the Maclaurin expansion cannot be anything but $0$. Therefore, since computing the first term of the sequence as it was printed gives $-\frac12$, $\frac12$ should be added to compensate, as you suspect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.