• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Infinite Surds. As we can see there are ten terms of this sequence where is the general term of the sequence when, is the first term of the sequence.

Extracts from this document...

Introduction

Math Portfolio

Infinite Surds

Introduction

a surd is an irrational number that can not be written as a fraction of two integers but can only be expressed using the root sign.

Bellow an example of an infinite surd:

image00.png

This surd can be turned into a set of particular numbers sequence:

image01.pngimage02.png1.414213562

image17.pngimage02.png1.553773974

image32.pngimage02.png1.598053182

image45.pngimage02.png1.611847754

a5image49.pngimage02.png1.616121207

image03.pngimage02.png1.617442799

image04.pngimage02.png1.617851291

image05.pngimage02.png1.617977531

image06.pngimage02.png1.618016542

image07.png


image08.png

As we can see there are ten terms of this sequence where image09.pngis the general term of the sequence whenimage10.png,image11.png is the first term of the sequence...Etc.

A formula has been defined for image12.pngin terms ofimage09.png:

image13.png(1)

image14.png

...read more.

Middle

 gets closer to a fixed value.

To investigate more about this fixed value we take this equation image18.png into consideration as n gets bigger.

n

an-an+1

1

- 0.13956

2

-0.04428

3

-0.01380

4

-0.00427

5

-0.00132

6

-0.00040

7

-0.00013

8

-0.00004

9

-0.00001

We can figure out from the table above that when n gets larger, the term (an+an+1) gets closer to zero but it never reaches it

So we can come to the conclusion:

When n approaches infinity, lim (an-an+1) →0

An expression can be obtained in the case of the relation between n and an to get the exact value of the infinite surd:

image19.png= x

If we apply formula (1) to this:

image20.pngimage21.png

x2=1+x   →   image22.png

The equation can be solved using the solution of a quadric equation:

image23.pngimage24.png

Whereas a=1, b=-1, c=-1

Two solutions for x were obtained:

x=1.618033989 and

...read more.

Conclusion

image23.pngimage24.png

The negative solution is ignored so:

image44.png

To find some values of k to make the expression an integer:

We can see that 4k is an even number and 4k+1 is odd, so image46.png is an odd number if 4k+1 is a perfect square hence 1+image46.png is an even number and possible to be divided by 2. As a result if 4k+1 is a perfect square we can obtain an integral number in the result.

For example let k=2:

image47.png

k=3

image48.png← not integer because13 is not a perfect square

Thus we come to the conclusion that only limited values of k can be used to make the result an integer and those values are any value of k can make 4k+1 a perfect square such as k = 2,6,12…etc.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    Y���K{��v^,����]Z�i-��" �o�_S'�+["&�v-�o(tm)F"����; ��"߳�i�&���Z? ^xCRk�'�Q�j-1i0x M �\G{��g��...���x6_^C�4�s��K#�uz�3/4 �2��I��I�7��N����"3/4�/�xfO ]_Eww�...���4 1/4�"[��"�B''y��e��Z-�u�(um6��]�~,-�?\ 1/2�^�H��c� ~e��(tm)�.-k�d1/4��:[�zIl��� �g]b��Aq�Cwm��"J�t�4�e�Þ�\x�[ׯm��[˨�[\x-�Z���o�Ec��}uiyeZ|+��f��� t��O��"�F^C�Z�{e���7�W�1/2]6��_M�-(c)owu���^�cv�!�:-{����޹��O� �}{k(c)]^k^0�m=m|-�[ nK;o�'�...4�-2 �u��}��g"T'(tm)� ;��/�d�c�/�_D[ jVv����'����doq��-��x��-��إ����Fy/(tm)��.��o�|S(r)�f�F"���{8u�...�N�(r)��7...k����V="�e�K×��7�Q�c[-��.��U��|I"�-~#Ô´MR�Þ�S�z���*:f(tm) x���^�K�Y<-) �"�a%IJ�iz5 (c)"�� "/X�7�9A�\jW�i�`���<M�]Óµñ¨­7KM2�i��5��Oo� J��?�o3/45+�1�"��kL��t�r� �F-�1/4=��:...���rZ�o�=��z"s1��,�k]2}HX�&����IQ-��@:��#����"h�M��Q^\�GX�<(c)Cek(c)jn}NmB�� MO��X�FKm:�� �'�@8�ó�|9�hZN��j-�z��{|c2[����J�(c)^˨�u� x���6��k�nt�(tm)�>)m�Õ� ��� �P"�4��I41/2_I:����K�7D��棡Y��uD���Æ{�ͤin�i6-�B-�@σ~|>�cq��w�:@���Û��ײ��5=3T�^��x5�y��."1/4�T������T1��l"��ҭ�ЭV0�!��c�t� ��X��s��à²Y�5)�"E�<;�����N-o�t+�� b"c(�)�-�+�|foh~*���&�|Cj"1/4��D7V�m�--(r)��CO :"V~&���J�\[إ��V��� '�o�� ���k&��.�Çtm!,��,-×Z��|Mi�h�'�Mw4�5�m�E��O���G��K�O�O��y/� �GH�o�x'�]Yx��W'�?\x��E-�VU��1/2_x'��zn�R����"ê0�r� 3/4 |����,<L|f�����1/2:-CP�X���g�:Ä{Qn���Z�i�]9�l�-MM?Î[{�h��( � ( �� ��"8x����_��5��(��G_�2�M�������U��@����'�*���}kS�>1/2sw�;D�е�}F�E1/2��t�.no,"�l�t(c)V -S�Iduy��-$K"�y���4|5���1/4E�g�-�;�b��qÜ")b�|I�+�$-R(tm)���u]V�IY5(tm)5mvvjq3/4���VW`׳_�M ]OPÕµ1/2RD�E��-Z��-�\�P�� �ⷵԭ�ê��GÇ"hW]:�Ú�-��Y�� ��v�M�...1/4�o�I1/4Cka���V��-z-�>1/2�&��w�-��B�H�A�'����k�He�.-Z�c���xo[��*�5m2q(c)�R NK��k�[���O�u�2,K�\Il���íµ"�<�U�M^���VÖ�jK�3�×u�l[��\�"J�f�"��E��Ù�(c)��M� ��j�m�j�F�a(c).�."�iò¶¢k~�q=�dj�����M��Ço1/4=�i�ς�T1�{Q-A���&��Ò��P�>i���oyg����Fk "y���"��.5;?x"�2x"_xb};UÖ´Û?B��<m4�9�4"yc����TzÊn��W*�1/2�Ø5��^���� X|&��"���;G�"�-��޳�i��rx^��\�"����I���7...��(tm)������H�1/4��-0 �-��~^�Y��-)�5�}6��Ú��ޡ���(r)��&��]?\���Ñ���-.�Q�_�2�'�-R�]X���� ���?�"�_ YM�ޭ��t��> i.t9�1/4�-�3/4��Zj��L"�ke(r)�CC_�#`N-���1/4w(��k�}V�v�i(r)\j�x��:��<E�]�uÜf'/�<)��P��eg��� Z�ZIlZ����-�T��V�g �߲ςu(�ψ�;�-�k�""j����\3/4�<5�Bo�t� ���^�m�i��m�� �OU�3/4-�]M��M�~Ï¿ �1w{�}s_3/4��1/47-(tm)u�^I�� �_����T��vhGT�����K�]_\Z��u1��Ù@�/�?�]�_�wOլ�uk�1/2�C֬����&�Hn4?

  2. This essay will examine theoretical and experimental probability in relation to the Korean card ...

    Starting with October, probability of winning with this hand is split into two different ways. First one is when player 2 has one card that has ? and second one is when player 2 does not have card with ?.

  1. Infinite Summation Internal Assessment ...

    This graph displays the line where =1 and =5 and the value of approaches the horizontal asymptote, which is 5. This creates a longer and steeper line compare to the previous graph since it reaches towards the horizontal asymptote of 5 in the same values of .

  2. Arithmetic Sequence Techniques

    = 32 2 = n (a4 + an-3) = 32 2 = n (8) = 32 2 = 8n = 64 Therefore: n = 8 3. If a1 + a2 + a3 = 5, an-2 + an-1 + an = 10 and Sn = 20, what is n?

  1. Infinite Summation - In this portfolio, I will determine the general sequence tn with ...

    2.997113 + = 2.999555 S7 = S6 + t7 = 2.999555 + = 2.999938 S8 = S7 + t8 = 2.999938 + = 2.999991 S9 = S8 + t9 = 2.999991 + = 2.999997 S10 = S9 + t10 = 2.999997 + = 2.999998 Now, using Microsoft Excel, I will

  2. Solution for finding the sum of an infinite sequence

    I got the same results. The method to do this is shown in the appendix. Because I got the same result, I will use Excel to graph. The method is shown in the Appendix. By looking at this graph, I can say that when value is past 10, the is 2 and remains as it is.

  1. Math Surds

    as all the values are positive and the trend suggests that they will remain so.

  2. In this portfolio, I will determine the general sequence tn with different values of ...

    4,968956 6 4,993095 7 4,998645 8 4,999761 9 4,999961 10 4,999993 From this graph we can see that as , the value of n increases as well, but it does not exceed 5, thus the greatest value of this relationship will be 5 and therefore the domain of this relationship is , as n approaches .

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work