Math 1502

Quiz #1  Jan 22 Tom Morley

Problem 1

M . Indurain needs to know the following : Does the series 1 + 1/2^3 + 1/3^3 + 1/4^3&n ... 62371; 1 + 1/2^3 + 1/3^3 + 1/4^3   + ... + 1/n^3   by using integrals .

Ans

Converges because

∫_1^Infinity1/x^3x

1/2

Is finite . OTOH, 1 + 1/2^3 + 1/3^3 + 1/4^3   + ... + 1/n^3    is less than

∫_1^(n + 1) 1/x^3x, and greater than 1 + ∫_1^n1/x^3x

Quiz #1  Jan 22   Tom Morley

Problem 2 (16 points)

Eddy  Merckx Needs find some series:

a) (2x^2)/(1 - x^3)

Start with 1/1 - t :

In[54]:=

s = Normal[Series[1/(1 - t), {t, 0, 10}]]

Out[54]=

1 + t + t^2 + t^3 + t^4 + t^5 + t^6 + t^7 + t^8 + t^9 + t^10

1 + t + t^2 + t^3 + t^4 + t^5 + t^6 + t^7 + t^8 + t^9 + t^10 + ... ..

Substitute t = -x^3

In[56]:=

s   = s /. t (-x^3) //Expand

Out[56]=

1 - x^3 + x^6 - x^9 + x^12 - x^15 + x^18 - x^21 + x^24 - x^27 + x^30

1 - x^3 + x^6 - x^9 + x^12 - x^15 + x^18 - x^21 + x^24 - x^27 + x^30 + ...

Multiply by 2x^2

In[57]:=

2 x^2 s //Expand

Out[57]=

2 x^2 - 2 x^5 + 2 x^8 - 2 x^11 + 2 x^14 - 2 x^17 + 2 x^20 - 2 x^23 + 2 x^26 - 2 x^29 + 2 x^32

2 x^2 - 2 x^5 + 2 x^8 - 2 x^11 + 2 x^14 - 2 x^17 + 2 x^20 - 2 x^23 + 2 x^26 - 2 x^29 + 2 x^32 + ...




b) ∫_0^xsin(t^3)/t t

In[62]:=

s = Normal[Series[Sin[x], {x, 0, 10}]]

Out[62]=

x - x^3/6 + x^5/120 - x^7/5040 + x^9/362880

x - x^3/6 + x^5/120 - x^7/5040 + x^9/362880 + ...

Substitute t^3 for x .

In[63]:=

s = s /. x (t^3) //Expand

Out[63]=

t^3 - t^9/6 + t^15/120 - t^21/5040 + t^27/362880

t^3 - t^9/6 + t^15/120 - t^21/5040 + t^27/362880 +...

Divide by t

In[64]:=

s/t //Expand

Out[64]=

t^2 - t^8/6 + t^14/120 - t^20/5040 + t^26/362880

t^2 - t^8/6 + t^14/120 - t^20/5040 + t^26/362880 +...

Integrate

In[65]:=

∫_0^xst

Out[65]=

x^4/4 - x^10/60 + x^16/1920 - x^22/110880 + x^28/10160640

x^4/4 - x^10/60 + x^16/1920 - x^22/110880 + x^28/10160640 +...

Quiz #1  Jan 22 Tom Morley

Problem 3: (10 Points)

Greg Lemond wants you to use series and error bounds to estimate e^(1/3)   to within  10^(-4)

The series of E^(1/3) is

1 + (1/3)/1 ! + (1/3)^2/2 ! + ... .

In[66]:=

Sum[((1/3)^k)/k !, {k, 0, n}]

RowBox[{Error estimate is M ((1/3)^(n + 1))/(n + 1) !, ,, , RowBox[{RowBox[{where, , M, , i ... E^t, , for, , t, , between, , 0, , and, , RowBox[{1, /, 3.}], , M}], =, 2 will work . }]}]

In[71]:=

ErrorBound[n_]    = N [(2 (1/3)^(n + 1)) . (n + 1) !]

Out[71]=

RowBox[{RowBox[{(, RowBox[{2., , RowBox[{3., ^, RowBox[{(, RowBox[{RowBox[{-, 1.}], -, RowBox ... , , n}]}], )}]}]}], )}], ., RowBox[{RowBox[{(, RowBox[{RowBox[{1., }], +, n}], )}], !}]}]

In[74]:=

Table[{n, ErrorBound[n]}, {n, 1, 10}]//TableForm

Out[74]//TableForm=

1 RowBox[{0.222222, ., 2.}]
2 RowBox[{0.0740741, ., 6.}]
3 RowBox[{0.0246914, ., 24.}]
4 RowBox[{0.00823045, ., 120.}]
5 RowBox[{0.00274348, ., 720.}]
6 RowBox[{0.000914495, ., 5040.}]
7 RowBox[{0.000304832, ., 40320.}]
8 RowBox[{0.000101611, ., 362880.}]
9 RowBox[{0.0000338702, ., 3.6288*10^6}]
10 RowBox[{0.0000112901, ., 3.99168*10^7}]

We see that n = 9 will work .

Created by Mathematica  (April 23, 2004)