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I wanted to expand the function $(x+2)^{x+2}$ around $x = -1$, that is, using

Series[(x + 2)^(x + 2), {x, -1, 2}]

and Mathematica returns the same expression. Why does this happen? The first term of a series expansion is simply $1^1 = 1$. However, I cannot even get this, from

Series[(x + 2)^(x + 2), {x, -1, 0}]

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This Series[(x + 2)^(x + 2), {x, x0, 2}] /. x0 -> -1 works. –  b.gatessucks Apr 16 '13 at 18:12
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Both expressions work here with Mathematica 8.0.1. Series[(x + 2)^(x + 2), {x, -1, 2}] gives 1 + (x+1) + (x+1)^2 + O[x+1]^3. And Series[(x + 2)^(x + 2), {x, -1, 0}] gives 1 + O[x+1]^1. Do you use a fresh kernel? –  partial81 Apr 16 '13 at 18:13
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Yes, fresh kernel. See the output png file in the new edit. This is with Mathematica 9. By the way, @b.gatessucks: your workaround gives errors if the series is about a singular point. In that case, the workaround requires two series expansions (once for the x0, and once more again for the proper value) –  TSGM Apr 16 '13 at 18:17
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Very odd. Translate it and it works, e.g., Series[(x + 3)^(x + 3), {x, -2, 2}] gives 1+(x+2)+(x+2)^2+O[x+2]^3. (v9.0.1) –  Michael E2 Apr 16 '13 at 19:13
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Ok, just as confirmation: Series[(x + 2)^(x + 2), {x, -1, 2}] gives 1 + (x+1) + (x+1)^2 + O[x+1]^3 with v8.0.1 but (2 + x)^(2 + x) with v9.0.0. Seems to be a bug in the new version. This brings me to two questions: 1. Where to report such a bug? 2. Does Wolfram provide updates so that one can work with a software which has less bugs? –  partial81 Apr 17 '13 at 8:21

1 Answer 1

For version 9

$Version

"9.0 for Mac OS X x86 (64-bit) (January 24, 2013)"

f[x_] = (x + 2)^(x + 2);

f2[x_] = (Series[(x + 2)^(x + 2), {x, a, 2}] // Normal) /. a -> -1

2 + x + (1 + x)^2

Plot[{f[x], f2[x]}, {x, -2.5, 0}]
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