1. The document describes an investigation of infinite sequences and calculating the sum of the first n terms as n approaches infinity. 2. The analysis shows that as n increases, the sum approaches the value of the parameter a, regardless of the values of a and x. However, larger values of a and x cause the sum to converge more slowly. 3. The general conclusion is that the sum of the infinite sequence (lna)n/n! as n approaches infinity equals the value of the parameter a.

1. Kohshi Gokita
March 28, 2010
IB Math SL. Per. 8
Internal Assessment
Infinite Summation
The infinite sequence below is considered
𝑡0 = 1 𝑡1 =
( 𝑥𝑙𝑛𝑎)01
1!
𝑡2 =
( 𝑥𝑙𝑛𝑎)2
2!
…………..
So, the general term of this sequence is expressed as follows;
𝑡 𝑛 =
( 𝑥𝑙𝑛𝑎) 𝑛
𝑛!
The sum Tn(a,x) of the first (n+1) terms of the above sequence is considered
𝑇𝑛(a ,x) = 𝑡0 + 𝑡1 + 𝑡2 + 𝑡3 + …………….+ 𝑡 𝑛
= ∑ 𝑡 𝑘
𝑛
𝑘=0
=∑
(xlna)k
𝑘!
𝑛
𝑘=0
Aim;
In this task, the sum of infinite sequence 𝑡 𝑛 is considered, hence, the value of
𝑇𝑛 (a, x) as a n approaches∞,
𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(𝑎, 𝑥)
would be calculated

2. I. In this part of investigation, the value of x is fixed on 1 and the
effect of change in the value of a over the value of 𝑻 𝒏(a, x) would
be observed by numerical simulation.
1. Consider the following sequence of terms where x=1 and a=2. And calculate the
sum 𝑻 𝒏 of the first n terms of the above sequence for 0≤n≤10.
1, +
( 𝑙𝑛2)
1
+
( 𝑙𝑛2)2
2∗1
+
( 𝑙𝑛2)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
2 1 0 1 1
1 0.693147 1.693147
2 0.240227 1.933374
3 0.055504 1.988878
4 0.009618 1.998496
5 0.001333 1.999829
6 0.000154 1.999983
7 1.53E-05 1.999999
8 1.32E-06 2
9 1.02E-07 2
10 7.05E-09 2
0≤n≤20:
a x n t T
2 1 0 1 1
0
0.5
1
1.5
2
2.5
0 2 4 6 8 10 12

3. 1 0.693147 1.693147
2 0.240227 1.933374
3 0.055504 1.988878
4 0.009618 1.998496
5 0.001333 1.999829
6 0.000154 1.999983
7 1.53E-05 1.999999
8 1.32E-06 2
9 1.02E-07 2
10 7.05E-09 2
11 4.45E-10 2
12 2.57E-11 2
13 1.37E-12 2
14 6.78E-14 2
15 3.13E-15 2
16 1.36E-16 2
17 5.53E-18 2
18 2.13E-19 2
19 7.77E-21 2
20 2.69E-22 2
*According to data, as n increases, sum of infinity approaches to the value of a. In this
case, the sum gets closed to 2.
0
0.5
1
1.5
2
2.5
0 5 10 15 20 25

4. 2. Consider the following sequence of terms where x=1 and a=3. And calculate the
sum 𝑻 𝒏 of the first n terms of the above sequence for 0≤n≤10.
1, +
( 𝑙𝑛3)
1
+
( 𝑙𝑛3)2
2∗1
+
( 𝑙𝑛3)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
3 1 0 1 1
1 1.098612 2.098612
2 0.603474 2.702087
3 0.220995 2.923082
4 0.060697 2.983779
5 0.013336 2.997115
6 0.002442 2.999557
7 0.000383 2.99994
8 5.26E-05 2.999993
9 6.42E-06 2.999999
10 7.06E-07 3
0≤n≤20:
a x n t T
3 1 0 1 1
1 1.098612 2.098612
2 0.603474 2.702087
3 0.220995 2.923082
0
0.5
1
1.5
2
2.5
3
3.5
0 2 4 6 8 10 12

5. 4 0.060697 2.983779
5 0.013336 2.997115
6 0.002442 2.999557
7 0.000383 2.99994
8 5.26E-05 2.999993
9 6.42E-06 2.999999
10 7.06E-07 3
11 7.05E-08 3
12 6.45E-09 3
13 5.45E-10 3
14 4.28E-11 3
15 3.13E-12 3
16 2.15E-13 3
17 1.39E-14 3
18 8.49E-16 3
19 4.91E-17 3
20 2.7E-18 3
*According to data, as n increases, sum of infinity approaches to the value of a. In this
case, the sum gets closed to 3.
3. Consider the following sequence of terms where x=1 and a=4. And calculate the
sum 𝑻 𝒏 of the first n terms of the above sequence for 0≤n≤10.
0
0.5
1
1.5
2
2.5
3
3.5
0 5 10 15 20 25

6. 1, +
( 𝑙𝑛4)
1
+
( 𝑙𝑛4)2
2∗1
+
( 𝑙𝑛4)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
4 1 0 1 1
1 1.386294 2.386294
2 0.960906 3.3472
3 0.444033 3.791233
4 0.15389 3.945123
5 0.042667 3.987791
6 0.009858 3.997649
7 0.001952 3.999601
8 0.000338 3.99994
9 5.21E-05 3.999992
10 7.22E-06 3.999999
0≤n≤20:
a x n t T
4 1 0 1 1
1 1.386294 2.386294
2 0.960906 3.3472
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 2 4 6 8 10 12
Series1

7. 3 0.444033 3.791233
4 0.15389 3.945123
5 0.042667 3.987791
6 0.009858 3.997649
7 0.001952 3.999601
8 0.000338 3.99994
9 5.21E-05 3.999992
10 7.22E-06 3.999999
11 9.1E-07 4
12 1.05E-07 4
13 1.12E-08 4
14 1.11E-09 4
15 1.03E-10 4
16 8.89E-12 4
17 7.25E-13 4
18 5.59E-14 4
19 4.08E-15 4
20 2.82E-16 4
*According to data, as n increases, sum of infinity approaches to the value of a. In this
case, the sum gets closed to 4.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 5 10 15 20 25

8. 4. Now, consider a general sequence where x=1. And calculate the sum 𝑻 𝒏 of the
first n terms of the above sequence for 0≤n≤10 for different values of a.
1, +
( 𝑙𝑛𝑎)
1
+
( 𝑙𝑛𝑎)2
2∗1
+
( 𝑙𝑛𝑎)3
3∗2∗1
+ ………..
If x=1 and a=10, the equation will be
1, +
( 𝑙𝑛10)
1
+
( 𝑙𝑛10)2
2∗1
+
( 𝑙𝑛10)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
10 1 0 1 1
1 2.302585 3.302585
2 2.650949 5.953534
3 2.034679 7.988213
4 1.171255 9.159468
5 0.539383 9.698851
6 0.206996 9.905847
7 0.068089 9.973936
8 0.019598 9.993534
9 0.005014 9.998548
10 0.001154 9.999702
0
2
4
6
8
10
12
0 2 4 6 8 10 12

9. 0≤n≤20:
a x n t T
10 1 0 1 1
1 2.302585 3.302585
2 2.650949 5.953534
3 2.034679 7.988213
4 1.171255 9.159468
5 0.539383 9.698851
6 0.206996 9.905847
7 0.068089 9.973936
8 0.019598 9.993534
9 0.005014 9.998548
10 0.001154 9.999702
11 0.000242 9.999944
12 4.64E-05 9.99999
13 8.21E-06 9.999998
14 1.35E-06 10
15 2.07E-07 10
16 2.98E-08 10
17 4.04E-09 10
18 5.17E-10 10
19 6.27E-11 10
20 7.21E-12 10

10. *According to data, as n increases, sum of infinity approaches to the value of a. In this
case, the sum gets closed to 10.
According to whole observation of experiment, it is proved that as n increases, the
sum approaches to value of a. In addition, as a increases, the converge speed gets slow
and slow. In cases of a=2 and x=1, sum becomes 2 when n is 8. However, in case of
a=10 and x=1, sum becomes 10 when n is 14. Ultimately, the general statement of the
infinite summation is
∑
(lna)n
𝑛!
∞
𝑛=0
= 𝑎
II. In this part of investigation, how the change in the positive value of
x as well as the positive value of a would affect the value of 𝑻 𝒏(a,
x) would beobserved by numerical simulation.
1. Consider the following sequence of terms where x=5 and a=2. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
(5𝑙𝑛2)
1
+
(5𝑙𝑛2)2
2∗1
+
(5𝑙𝑛2)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
2 5 0 1 1
1 3.465736 4.465736
2 6.005663 10.4714
0
2
4
6
8
10
12
0 5 10 15 20 25

11. 3 6.938014 17.40941
4 6.011331 23.42074
5 4.166737 27.58748
6 2.406802 29.99428
7 1.19162 31.1859
8 0.51623 31.70213
9 0.198791 31.90092
10 0.068896 31.96982
0≤n≤20;
a x n t T
2 5 0 1 1
1 3.465736 4.465736
2 6.005663 10.4714
3 6.938014 17.40941
4 6.011331 23.42074
5 4.166737 27.58748
6 2.406802 29.99428
7 1.19162 31.1859
8 0.51623 31.70213
9 0.198791 31.90092
10 0.068896 31.96982
11 0.021707 31.99152
0
5
10
15
20
25
30
35
0 2 4 6 8 10 12

12. 12 0.006269 31.99779
13 0.001671 31.99946
14 0.000414 31.99988
15 9.56E-05 31.99997
16 2.07E-05 31.99999
17 4.22E-06 32
18 8.13E-07 32
19 1.48E-07 32
20 2.57E-08 32
*According to data, the graph’s shape changes. In this case, sum becomes 32 when n is
16. In addition, in the graph, there is a inflection point. The curve of the graph changes.
a=2, x=5  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(2,5) = 25
= 32
2. Consider the following sequence of terms where x=3 and a=2. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
(3𝑙𝑛2)
1
+
(3𝑙𝑛2)2
2∗1
+
(3𝑙𝑛2)3
3∗2∗1
+ ………..
0≤n≤10:
a x n t T
0
5
10
15
20
25
30
35
0 5 10 15 20 25

13. 2 3 0 1 1
1 2.079442 3.079442
2 2.162039 5.24148
3 1.498611 6.740091
4 0.779068 7.519159
5 0.324005 7.843165
6 0.112292 7.955457
7 0.033358 7.988814
8 0.008671 7.997485
9 0.002003 7.999488
10 0.000417 7.999905
0≤n≤20:
a x n t T
2 3 0 1 1
1 2.079442 3.079442
2 2.162039 5.24148
3 1.498611 6.740091
4 0.779068 7.519159
5 0.324005 7.843165
6 0.112292 7.955457
7 0.033358 7.988814
8 0.008671 7.997485
0
1
2
3
4
5
6
7
8
9
0 2 4 6 8 10 12

14. 9 0.002003 7.999488
10 0.000417 7.999905
11 7.88E-05 7.999984
12 1.36E-05 7.999997
13 2.18E-06 8
14 3.24E-07 8
15 4.49E-08 8
16 5.84E-09 8
17 7.15E-10 8
18 8.25E-11 8
19 9.03E-12 8
20 9.39E-13 8
*According to data, the graph’s shape changes. In this case, sum becomes 8 when n is
about5. In addition, in the graph, there is a inflection point. The curve of the graph
changes.
a=2, x=3  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(2,3) = 23
= 8
0
1
2
3
4
5
6
7
8
9
0 5 10 15 20 25

15. According to whole observation of investigation, it is proved that if the value of x
changes, the sum also changes. However, it is not that the sum approaches value of a.
The sum approaches the number of 𝑎 𝑥
. In investigation in part I, the sum becomes the
same value as a but it is that because x is fixed on 1. In the case of a=2 and x=5, the
total sum becomes 25
, 32. Ultimately, the general statement of the infinite summation
is
∑
(lna)n
𝑛!
∞
𝑛=0
= ax
III. In this part of investigation, how the negative value of x as well as
the positive value of a would affect the value of 𝑻 𝒏(a, x) would be
observed by numerical simulation. In addition, how the change in
the positive value of a as well as the value of x would affect the
valueof 𝑻 𝒏(a, x) would beobserved by numerical simulation.
1. Consider the following sequence of terms where x= -2 and a=2. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
(−2𝑙𝑛2)
1
+
(−2𝑙𝑛2)2
2∗1
+
(−2𝑙𝑛2)3
3∗2∗1
+ ………..
0≤ 𝑛 ≤ 10
a x n t T
2 -2 0 1 1
1 -1.38629-0.38629
2 0.9609060.574612
3 -0.444030.130579
4 0.153890.284469
5 -0.042670.241801
6 0.009858 0.25166
7 -0.001950.249707
8 0.0003380.250046
9 -5.2E-050.249994
10 7.22E-060.250001

16. 0≤ 𝑛 ≤ 20
a x n t T
2 -2 0 1 1
1 -1.38629 -0.38629
2 0.960906 0.574612
3 -0.44403 0.130579
4 0.15389 0.284469
5 -0.04267 0.241801
6 0.009858 0.25166
7 -0.00195 0.249707
8 0.000338 0.250046
9 -5.2E-05 0.249994
10 7.22E-06 0.250001
11 -9.1E-07 0.25
12 1.05E-07 0.25
13 -1.1E-08 0.25
14 1.11E-09 0.25
15 -1E-10 0.25
16 8.89E-12 0.25
17 -7.3E-13 0.25
18 5.59E-14 0.25
19 -4.1E-15 0.25
20 2.82E-16 0.25
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
Series1

17. *According to the graph, the shape has dramatically changed from the graph of 0≤ 𝑥.
When n is 2, the sum value becomes negative but when n is 3, the value goes to positive
again. However, when n is 4, the sum value does not go to negative but it decreases
once again. From n becomes 4, the sum value constantly becomes 0.25.
a=2, x=-2  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(2,−2) = 2−2
= 0.25
2. Consider the following sequence of terms where x= -2 and a= 1/2. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
(−2ln(
1
2
))
1
+
(−2ln(
1
2
))
2
2∗1
+
(−2ln(
1
2
))
3
3∗2∗1
+ ………..
0≤n≤10
a x n t T
0.5 -2 0 1 1
11.3862942.386294
20.960906 3.3472
30.4440333.791233
4 0.153893.945123
50.0426673.987791
60.0098583.997649
70.0019523.999601
80.000338 3.99994
95.21E-053.999992
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Series1

18. 107.22E-063.999999
0≤n≤20
a x n t T
2 -2 0 1 1
1 -1.38629 -0.38629
2 0.960906 0.574612
3 -0.44403 0.130579
4 0.15389 0.284469
5 -0.04267 0.241801
6 0.009858 0.25166
7 -0.00195 0.249707
8 0.000338 0.250046
9 -5.2E-05 0.249994
10 7.22E-06 0.250001
11 -9.1E-07 0.25
12 1.05E-07 0.25
13 -1.1E-08 0.25
14 1.11E-09 0.25
15 -1E-10 0.25
16 8.89E-12 0.25
17 -7.3E-13 0.25
18 5.59E-14 0.25
19 -4.1E-15 0.25
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 2 4 6 8 10 12
Series1

19. 20 2.82E-16 0.25
*According to the graph, the shape becomes back to the first type of graph. Although
the x is negative number, the graph shapes same looking as the graph of 𝑇𝑛(𝑎,1 ). It
might be because of a is half of 1.
a=1/2, x= -2  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(
1
2
, −2) = (
1
2
)−2
= 4
3. Consider the following sequence of terms where x=2 and a=1/2. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
(2ln(
1
2
))
1
+
(2ln(
1
2
))
2
2∗1
+
(2ln(
1
2
))
3
3∗2∗1
+ ………..
0≤n≤10
a x n t T
0.5 2 0 1 1
1-1.38629-0.38629
20.9609060.574612
3-0.444030.130579
4 0.153890.284469
5-0.042670.241801
60.009858 0.25166
7-0.001950.249707
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 5 10 15 20 25
Series1

20. 80.0003380.250046
9-5.2E-050.249994
107.22E-060.250001
0≤n≤20
a x n t T
0.5 2 0 1 1
1 -1.38629-0.38629
2 0.9609060.574612
3 -0.444030.130579
4 0.153890.284469
5 -0.042670.241801
6 0.009858 0.25166
7 -0.001950.249707
8 0.0003380.250046
9 -5.2E-050.249994
10 7.22E-060.250001
11 -9.1E-07 0.25
12 1.05E-07 0.25
13 -1.1E-08 0.25
14 1.11E-09 0.25
15 -1E-10 0.25
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
Series1

21. 16 8.89E-12 0.25
17 -7.3E-13 0.25
18 5.59E-14 0.25
19 -4.1E-15 0.25
20 2.82E-16 0.25
*According to graph, it becomes exactly same as the graph of 𝑇𝑛(2,−2 ) although this
graph’s x value is not negative. This graph is 𝑇𝑛(
1
2
, 2 ), but it becomes exact same
looking graph. When n is 2, the sum value becomes negative but when n is 3, the value
goes to positive again. However, when n is 4, the sum value does not go to negative but
it decreases once again. From n becomes 4, the sum value constantly becomes 0.25.
a=1/2, x=2  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(
1
2
, 2) =
1
2
2
= 0.25
 According to whole observation of investigation, it is proved that the sum value
follows the general statement of ∑
(lna)n
𝑛!
∞
𝑛=0 = ax
although value of x and a become
rational.
IV. In this part of investigation, how the rational value of x as well as
the rational value of a such as 1/3, would affect the value of 𝑻 𝒏(a,
x) would be observed by numerical simulation. In addition, how the
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Series1

22. irrational value of x as well as the irrational value of x such as 𝛑
or √𝟑 would affect the value of 𝑻 𝒏(a, x) would be observed by
numerical simulation.
1. Consider the following sequence of terms where x=1/3, and a=1/3. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10 and 0≤n≤20.
1, +
(
1
3
ln(
1
3
))
1
+
(
1
3
ln(
1
3
))
2
2∗1
+
(
1
3
ln(
1
23
))
3
3∗2∗1
+ ………..
0≤n≤10
a x n t T
0.333333 0.333333 0 1 1
1 -0.3662 0.633796
2 0.067053 0.700849
3 -0.00818 0.692664
4 0.000749 0.693413
5 -5.5E-05 0.693358
6 3.35E-06 0.693361
7 -1.8E-07 0.693361
8 8.02E-09 0.693361
9 -3.3E-10 0.693361
10 1.2E-11 0.693361
0≤n≤20
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
Series1

23. a x n t T
0.333333 0.333333 0 1 1
1 -0.3662 0.633796
2 0.067053 0.700849
3 -0.00818 0.692664
4 0.000749 0.693413
5 -5.5E-05 0.693358
6 3.35E-06 0.693361
7 -1.8E-07 0.693361
8 8.02E-09 0.693361
9 -3.3E-10 0.693361
10 1.2E-11 0.693361
11 -4E-13 0.693361
12 1.21E-14 0.693361
13 -3.4E-16 0.693361
14 8.95E-18 0.693361
15 -2.2E-19 0.693361
16 5E-21 0.693361
17 -1.1E-22 0.693361
18 2.19E-24 0.693361
19 -4.2E-26 0.693361
20 7.73E-28 0.693361
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Series1

24. *According to graph, it also shapes like 𝑇𝑛(2,−2 ) and 𝑇𝑛(
1
2
, 2 ). But the difference is
that sum value never goes to negative although above both of sum value goes to
negative at least once. And from n=5, the sum value constantly get 0.693361.
a=1/3, x=1/3  𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(
1
3
,
1
3
) =
1
3
1
3
= 0.693361.
2. Consider the following sequence of terms where x=√3, and a=π. And calculate the
sum 𝑇𝑛 of the first n terms of the above sequence for 0≤n≤10.
1, +
((√3)ln(𝜋))
1
+
((√3)ln(𝜋))
2
2∗1
+
((√3)ln(𝜋))
3
3∗2∗1
+ ………..
0≤n≤10
a x n t T
𝜋 √3 0 1 1
1 1.98273 2.98273
2 1.96561 4.94834
3 1.299091 6.247431
4 0.643937 6.891368
5 0.255351 7.146719
6 0.084382 7.231101
7 0.023901 7.255002
8 0.005924 7.260925
9 0.001305 7.26223
10 0.000259 7.262489

25. 0≤n≤20
a x n t T
3.141593 1.732051 0 1 1
1 1.98273 2.98273
2 1.96561 4.94834
3 1.299091 6.247431
4 0.643937 6.891368
5 0.255351 7.146719
6 0.084382 7.231101
7 0.023901 7.255002
8 0.005924 7.260925
9 0.001305 7.26223
10 0.000259 7.262489
11 4.66E-05 7.262536
12 7.71E-06 7.262543
13 1.18E-06 7.262545
14 1.66E-07 7.262545
15 2.2E-08 7.262545
16 2.73E-09 7.262545
17 3.18E-10 7.262545
18 3.5E-11 7.262545
19 3.66E-12 7.262545
20 3.62E-13 7.262545
0
1
2
3
4
5
6
7
8
0 2 4 6 8 10 12
Series1

26. *According to the graph, it becomes same shape as the graph of 𝑇𝑛(2,1). When n
becomes 5, the sum value begins to constantly get 7.262545.
According to whole observation of investigation, it is proved that although value of x
and becomes something irrational, sum value can be defined by the general statement of
∑
(lna)n
𝑛!
∞
𝑛=0
= 𝑎
Conclusion:
According to all the observation of data, it can be defined that all of sum value follows
that general statement above. However, the shapes of graphing are not always same.
Some differs from others. From all the data that gathered, sum value of xlna can be
defined that
x a lna xlna
x>0 a>1 lna>0 Positive +
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25
Series1
0
10
20
30
40
0 5 10 15

27. x>0 0 <
𝑎 <1
ln𝑎 <0 Negative
x<0 a>1 lna>0 Negative
x<0 0 <
𝑎 <1
ln𝑎 <0 Positive +
Ultimately, to answer to the aim, whatever the a and x are, infinite sum value follows
the general statement as n increases. So, 𝑡 𝑛 is considered, hence, the value of 𝑇𝑛 (a, x)
as a n approaches∞,
𝑙𝑖𝑚
𝑛→∞
𝑇𝑛(𝑎, 𝑥)
It can be conclude with
aa>0
any positive real numbers.
xany real numbers.
-0.5
0
0.5
1
1.5
0 10 20
Series1
-0.5
0
0.5
1
1.5
0 5 10 15
Series1
0
2
4
6
0 5 10 15
Series1