Mathematica - 値と指数関数を使って配列をプロットする

Question

mathematicaについて質問があります。私はtempDepと呼ばれる値を持つ配列があります。

{10.7072,11.5416,12.2065,12.774,13.2768,13.7328,14.1526,14.5436,14.9107,15.2577,15.5874,15.9022,16.2037,16.4934,16.7727,17.0425,17.3036,17.5569,17.803,18.0424,18.2756,18.503,18.725,18.9419,19.154,19.3615,19.5647,19.7637,19.9588,20.1501,20.3378,20.5219,20.7025,20.8799,21.0541,21.2252,21.3933,21.5584,21.7207,21.8801,22.0368,22.1908,22.3423,22.4911,22.6374,22.7813,22.9228,23.0619,23.1987,23.3332,23.4655,23.5955,23.7235,23.8493,23.973,24.0947,24.2143,24.332,24.4478,24.5616,24.6736,24.7837,24.892,24.9986,25.1034,25.2064,25.3078,25.4075,25.5055,25.602,25.6968,25.7901,25.8819,25.9722,26.061,26.1483,26.2342,26.3186,26.4017,26.4835,26.5638,26.6429,26.7207,26.7972,26.8724,26.9464,27.0192,27.0908,27.1612,27.2304,27.2986,27.3656,27.4315,27.4963,27.56,27.6227,27.6844,27.7451,27.8048,27.8635,27.9212,27.978,28.0338,28.0887,28.1428,28.1959,28.2482,28.2996,28.3502,28.3999,28.4488,28.497,28.5443,28.5908,28.6366,28.6817,28.726,28.7695,28.8124,28.8545,28.896,28.9368,28.9769,29.0163,29.0551,29.0933,29.1308,29.1678,29.2041,29.2398,29.2749,29.3095,29.3435,29.3769,29.4098,29.4421,29.474,29.5053,29.536,29.5663,29.5961,29.6254,29.6542,29.6825,29.7104,29.7378,29.7647,29.7913,29.8173,29.843,29.8682,29.893,29.9175,29.9415,29.9651,29.9883,30.0112,30.0336,30.0557,30.0775,30.0989,30.1199,30.1406,30.1609,30.1809,30.2006,30.22,30.239,30.2578,30.2762,30.2943,30.3121,30.3297,30.3469,30.3639,30.3806,30.397,30.4131,30.429,30.4446,30.4599,30.4751,30.4899,30.5045,30.5189,30.533,30.5469,30.5606,30.5741,30.5873,30.6003,30.6131,30.6257,30.6381,30.6503,30.6623,30.674,30.6856} 

をし、私が何をしたいのか

ListPlot[tempDep] 

を使用して、それをプロットしています指数と一緒にこのプロットを表示することです（つまりはかなりになりますこのlistPlotと同じ）を1つのグラフに表示します。このplzで私を助けることができますか？

Answer 1

でこのような何かを機能を補間のグラフをプロットすることができますか？

data = Table[Sin[x], {x, 0, 2 Pi, 0.3}]; 

Show[ 
ListPlot[data, PlotStyle -> PointSize[0.02]], 
ListLinePlot[data, 
    InterpolationOrder -> 2, 
    PlotStyle -> Directive[Thick, Orange]] 
] 

Answer 2

あなたは

f = Interpolate[tempDep] 

を使用して、おそらく

Plot[f,{x,1,198}] 

Answer 3

あなたのデータは他の何かに従うが、あなたは指数関数的にフィットしたい場合ように私には思える：

model = a + b Exp[c + d x]; 
tempDep1 = Partition[Riffle[Range@Length@tempDep, tempDep], 2]; 
fit  = FindFit[tempDep1, model, {a, b, c, d}, x, Method -> NMinimize]; 
modelf = Function[{x}, Evaluate[model /. fit]] 
Plot[modelf[t], {t, 0, Length@tempDep}, Epilog -> Point@tempDep1]