MHB How Many Real Roots Does This Equation Have?

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The equation presented is a finite sum of the exponential series, specifically the Taylor series for \( e^x \) truncated at the sixth term. The discussion centers on determining the number of real roots of this equation, which involves analyzing the behavior of the function defined by the series. Members contributed solutions, with lfdahl and Opalg recognized for their correct answers. The nature of the series suggests that it is always positive for real values of \( x \), indicating that the equation has no real roots. Therefore, the conclusion is that the equation has zero real roots.
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Here is this week's POTW:

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Find the number of real roots of the equation $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}=0$.

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Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. Opalg

Solution from Opalg:
Let $P(x) = 1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}$. Then
$$\begin{aligned}720P(x) &= x^6 + 6x^5 + 30x^4 + 120x^3 + 360x^2 + 720x + 720 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9x^4 + 54x^3 + 234x^2 + 540x + 495 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^4 + 6x^3 + 26x^2 + 60x + 55) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9x^2 + 36x + 39\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9(x + 2)^2 + 3\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^2 + 3x + 4)^2 + 81(x + 2)^2 + 27.\end{aligned}$$ That is a sum of squares, so it is always positive (with minimum value at least $27$). Therefore $P(x)$ is always positive and so the equation $P(x) = 0$ has no real roots.
 
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