Infinite equation

Calculus Level 5

$\mathscr{E} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} +\cdots$

For how many of the following value(s) of $\mathscr{E}$ will there be a real number value of $x$ that satisfies the above equation?

$-2, \ -\dfrac{3}{2}, \ -1, \ -\dfrac{1}{2}, \ 0, \ \dfrac{1}{2}, \ 1, \ \dfrac{3}{2}, \ 2$

6 4 3 2 8 7 5 9

1 solution

Chew-Seong Cheong
Jun 25, 2016

Relevant wiki: Maclaurin Series

$\mathscr{E} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$ is the Maclaurin series for $\ln(1+x)$ for $-1 < x \le 1$ or $0 < 1+x \le 2$ and from the list are $- \frac 12$ , $0$ , $\frac 12$ , $1$ , $\frac 32$ , and $2$ , altogether $\boxed{6}$ of them.

I think you mean that $E \in ( - \infty, \ln 2 ]$ ?

Calvin Lin Staff - 4 years, 11 months ago

So, the solution is wrong. The problem is not properly worded.

Chew-Seong Cheong - 4 years, 11 months ago

No there are $\boxed 6$ values of $\mathscr E$ that belong to $(-\infty,\ln 2]$ i.e $-2,-1.5,-1,-0-5,0,0.5$ ... I guess you need some edits in your soln.

Rishabh Jain - 4 years, 11 months ago

@Rishabh Jain – Thanks, yes, you are right. I have changed the solution.

Chew-Seong Cheong - 4 years, 11 months ago

How can x=3/2 and 2 when x cannot be more than 1? (Unless I misinterpreted your answer)

Curtis Clement - 4 years, 11 months ago

Thanks, they are meant for $\mathscr E$ .

Chew-Seong Cheong - 4 years, 11 months ago

Infinite equation

1 solution

0 pending reports