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This is a follow-up question of the previous question I asked. The previous question proved that the root of $\sum_{k=0}^n\frac{x^k}{k!}=\frac{e^x}{2}$ lies in the interval $(n,n+1)$. However, while I am playing with geogebra, I found that the root is close to $n+\frac{2}{3}$. I have tried to deal it with python for larger value, and it looks like $n+2/3+\frac{0.01975^+}{n}+o(1/n)$. I tried to imitate the solution in the the previous question to prove this observation to prove this, but failed... Can someone prove or give a hint on how to prove this observation?

P.S. I have read the post how to ask a good question. I really don't know how to go on with this problem. Please don't regard this question as no-clue questions, thanks you really much.

P.P.S. Here attaches the value $n(r-n-\frac 23)$, where $r$ is the root, for $n$ from $0$ to $10^5$ where $n$ is a multiple of $1000$. The python code is as follows (it used the upper gamma incomplete function, as a solution of the previous question suggested):

import scipy.special as spec
def find_root(n):
    root = n
    for i in range(100):
        step = 0.5**i
        if 2*spec.gammaincc(n+1, root)>1:
            root+=step
        else: root-=step
    return root

for i in range(101):
    print(i1000, (find_root(i1000)-i1000-2/3)1000*i)

And the results are as follows:

0 0.0
1000 0.019740550745495078
2000 0.019746816785071175
3000 0.019748905288108354
4000 0.0197499515102173
5000 0.019750574817289923
6000 0.019751001673506963
7000 0.019751297562486947
8000 0.019751513415933175
9000 0.019751703803527754
10000 0.01975183598346142
11000 0.019751942697543434
12000 0.019752056687583064
13000 0.01975209791804655
14000 0.019752197356170953
15000 0.019752224034719212
16000 0.019752323472843614
17000 0.01975234287543426
18000 0.019752435037601046
19000 0.01975241806040362
20000 0.01975243018703665
21000 0.019752536901118667
22000 0.019752556303709312
23000 0.01975258298225757
24000 0.019752595108890603
25000 0.019752527199989878
26000 0.019752633914071893
27000 0.01975267514453538
28000 0.019752672719253184
29000 0.019752743053547128
30000 0.019752624212943104
31000 0.019752650891491363
32000 0.019752808537276678
33000 0.01975276973220641
34000 0.01975267271947523
35000 0.019752502947167905
36000 0.019752653316995605
37000 0.01975268727150148
38000 0.01975292495282055
39000 0.01975288614775028
40000 0.01975296375800184
41000 0.01975277943377929
42000 0.019752944355522217
43000 0.019752701823638752
44000 0.019752983160703508
45000 0.019752696973074357
46000 0.019752527200767034
47000 0.019753099576136357
48000 0.01975288614808335
49000 0.019752978310250136
50000 0.01975301226475601
51000 0.019752507798398433
52000 0.01975313838142867
53000 0.019752779434223378
54000 0.01975300256362722
55000 0.019753036518133094
56000 0.019752808538164857
57000 0.01975301711565347
58000 0.019752672720363407
59000 0.01975324509573273
60000 0.01975276973320561
61000 0.0197526581685592
62000 0.019752575707743247
63000 0.019753046219705972
64000 0.019752730928246365
65000 0.01975301226531112
66000 0.019753410017697703
67000 0.019752410786222363
68000 0.019752342877321638
69000 0.01975364284845238
70000 0.019752672720807496
71000 0.019753827172785954
72000 0.019753351810258835
73000 0.01975363314732359
74000 0.019752866746491904
75000 0.01975349732952214
76000 0.01975366225126507
77000 0.0197536234461948
78000 0.019753642848785447
79000 0.019753982393511116
80000 0.01975257570840938
81000 0.019751954826729978
82000 0.019752381682947018
83000 0.01975411821153461
84000 0.019752536903450135
85000 0.019752672721362607
86000 0.01975228467032686
87000 0.019754079406575364
88000 0.01975319659042185
89000 0.019752313774268337
90000 0.01975422492583867
91000 0.019753895082463835
92000 0.01975408910803722
93000 0.01975227496930909
94000 0.019754011497896684
95000 0.01975403090048733
96000 0.019752420488572398
97000 0.019752090645197562
98000 0.019753215993345563
99000 0.019753089876783925

Also, when I substitute the $e^x/2$ into other ratios (say, $e^x/3$), the root and $n$ is starting to differ more than a constant...

JetfiRex
  • 2,451
  • root of a strictly positive function? – Luciano Feb 04 '22 at 23:00
  • @Luciano I mean, root, not roots... we only consider one specific root of $\sum_{k=0}^n\frac{x^k}{k!}=\frac{e^x}{2}$. – JetfiRex Feb 04 '22 at 23:34
  • 1
    This is equivalent to the asymptotic inversion with respect to $x$ and for large $n$ the normalised incomplete gamma function $Q$: $$ Q(n + 1,x) = \frac{1}{2}. $$ From this result, we have $$ x = n + \frac{2}{3} + \frac{8}{{405(n + 1)}} + \mathcal{O}!\left( {\frac{1}{{(n + 1)^2 }}} \right) = n + \frac{2}{3} + \frac{8}{{405n}} + \mathcal{O}!\left( {\frac{1}{{n^2 }}} \right). $$ See this paper for proofs. – Gary Feb 04 '22 at 23:35
  • @Gary Wow, thanks for the comment (this is actually stronger than the previous question)! the $8/405$ does match the $0.01975$ there! – JetfiRex Feb 04 '22 at 23:39
  • Somehow connected : https://math.stackexchange.com/q/3883132 – Jean Marie Feb 05 '22 at 00:34
  • Connected https://math.stackexchange.com/a/1866453/305862 (one of your answers @Claude Leibovici ...) The present question and its answer should interest you, Claude. – Jean Marie Feb 05 '22 at 12:49

1 Answers1

4

This is equivalent to the asymptotic inversion with respect to $x$ and for large $n$ the normalised incomplete gamma function $Q$: $$ Q(n + 1,x) = \frac{1}{2}. $$ From this result, we have $$ x = n + \frac{2}{3} + \frac{8}{{405(n + 1)}} + \mathcal{O}\!\left( {\frac{1}{{(n + 1)^2 }}} \right) = n + \frac{2}{3} + \frac{8}{{405n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ The natural large parameter is $(n+1)$ but you can re-expand in $n$. In particular, $$ x \sim n + \frac{2}{3} + \frac{8}{{405n}} - \frac{{64}}{{5103n^2 }} + \frac{{2944}}{{492075n^3 }} + \ldots \,. $$

For the asymptotic inversion of the normalised incomplete gamma function in general, consider this paper. For this special case, see this paper.

Gary
  • 31,845