2D And 3D Sequences Project Essay Research

2D And 3D Sequences Project Essay, Research Paper

Plan of Investigation In this experiment I am traveling to necessitate the followers: A reckoner A pencil A pen Variety of beginnings of information Paper Ruler In this probe I have been asked to happen out how many squares would be needed to do up a certain form harmonizing to its sequence. The form is shown on the front page. In this probe I hope to happen a expression which could be used to happen out the figure of squares needed to construct the form at any sequencial place. Firstly I will interrupt the job down into simple stairss to get down with and travel into more item to explicate my solutions. I will exemplify to the full any methods I should utilize and explicate how I applied them to this certain job. I will foremost transport out this experiment on a 2D form and so widen my probe to 3D. The Number of Squares in Each Sequence I have achieved the following information by pulling out the form and widening upon it. Seq. no.12345678 No. Of cubes151325416185113 I am traveling to utilize this following method to see if I can work out some kind of form: SequenceCalculationsAnswer 1=11 22 ( 1 ) +35 32 ( 1+3 ) +513 42 ( 1+3+5 ) +725 52 ( 1+3+5+7 ) +941 62 ( 1+3+5+7+9 ) +1161 72 ( 1+3+5+7+9+11 ) +1385 82 ( 1+3+5+7+9+11+13 ) +15113 92 ( 1+3+5+7+9+11+13+15 ) +17 145 What I am making above is shown with the assistance of a diagram below ; If we take sequence 3: 2 ( 1+3 ) +5=13 2 ( 1 squares ) 2 ( 3 squares ) 1 ( 5 squares ) The Patterns I Have Noticied in Transporting Out the Previous Method I have now carried out ny first probe into the form and have seen a figure of different forms. First I can see that the figure of squares in each form is an uneven figure. Second I can see that the figure of squares in the form can be found out by taking the uneven Numberss from 1 onwards and adding them up ( harmonizing to the sequence ) . We so take the summing up ( & # 229 ; ) of these uneven Numberss and multiply them by two. After making this we add on the following back-to-back uneven figure to the twofold sum. I have besides noticied something through the drawings I have made of the forms. If we look at the symetrical sides of the form and add up the figure of squares we achieve a square figure. Trying to Obtain a Formula Through the Use of the Difference Method I will now use Jean Holderness & # 8217 ; difference method to seek and happen a expression. Pos.in seq. 123456 No.of squar. ( degree Celsius ) 1513254161 1st differ. ( a+b ) 048121620 2nd differ. ( 2a ) 44444 We can now utilize the equation an & lt ; sup & gt ; 2 & lt ; /sup & gt ; + bn + c & # 8216 ; n & # 8217 ; bespeaking the place in the sequence. If a = 2 so c = 1 and a + B = 0 If 2 is equal to b- so b = -2 I will now work out the equation utilizing the information I have obtained through utilizing the difference method: 1 ) 2 ( n -1 ) ( n & # 8211 ; 1 ) + 2n & # 8211 ; 1 2 ) 2 ( n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 ) + 2n & # 8211 ; 1 3 ) 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 4n + 2 + 2n & # 8211 ; 1 4 ) 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 Therefore my concluding equation is: 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 Proving My Equation and Using it to Find the Number of Squares in Higher Sequences I will now turn out my equation by using it to a figure of sequences and higher sequences I have non yet explored. Sequence 3: 1. 2 ( 3 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 6 + 1 2. 2 ( 9 ) & # 8211 ; 6 + 1 3. 18 -5 4. = 13 The expression when applied to sequence 3 appears to be successful. Sequence 5: 1. 2 ( 5 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 10 + 1 2. 2 ( 25 ) & # 8211 ; 10 + 1 3. 50 & # 8211 ; 10 + 1 4. 50 & # 8211 ; 9 5. = 41 Successful Sequence 6: 1. 2 ( 6 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 12 + 1 2. 2 ( 36 ) & # 8211 ; 12 +1 3. 72 & # 8211 ; 12 + 1 4. 72 & # 8211 ; 11 5. = 61 Successful Sequence 8: 1. 2 ( 8 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 16 + 1 2. 2 ( 64 ) & # 8211 ; 16 + 1 3. 128 & # 8211 ; 16 + 1 4. 128 & # 8211 ; 15 5. = 113 Successful The expression I found seems to be successful as I have shown on the old page. I will now utilize the expression to happen the figure of squares in a higher sequence. So now I wil use the expression 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 to seek and happen the figure of squares contained in sequence 20. Sequence 20: 2 ( 20 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 40 + 1 2 ( 400 ) & # 8211 ; 40 + 1 800 & # 8211 ; 40 + 1 800 & # 8211 ; 49 = 761 Alternatively of exemplifying the form I am traveling to utilize the method I used at the start of this piece of coursework. The method in which Iused to look for any forms in the sequences. I will utilize this to turn out the figure of squares given by the equation is right. As shown below: 2 ( 1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37 ) + 39 = 761 I feel this proves the equation to the full. Using the Difference Method to Find an Equation to Establish the Number of Squares in a 3D Version of the Pattern Pos.in seq.012345 No.of squar.-1172563129 1st differ.26183866 2nd differ.412202836 3rd differ.8888 So hence

we get the equation ; an? + bn²+ cn + vitamin D We already know the values of ‘n’ ( place in sequence ) in the equation so now we have to happen out the values of a, B, degree Celsius, and d. If n = 0 so d = -1 and if n = 1 so d = 1 I can now acquire rid of vitamin D from the equation to do it easier to happen the remainder of the values. I will will take n = 2 to make this in the undermentioned manner: 1st computation _ 8a + 4b + 2c + d a + B + c +d 7a + 3b + degree Celsius D will ever be added to each side of the equation. 2nd 8a + 4b + 2c = 8 = 4a + 2b + degree Celsiuss = 4 2 So so n = 2 8a + 4b + 2c = 8 = 4a + 2b + degree Celsiuss = 4 n = 3 27a + 9b + 3c = 26 n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 16 4 To acquire rid of ‘c’ I will utilize this computation ; _16a + 4b + c = 16 4a + 2b + degree Celsiuss = 4 12a + 2b = 12 We can simplify this equation to: 6a + B = 6 My following computation is below: N =3_27a + 9b + 3c = 26 12a + 6b + 3c = 12 15a + 3b =14 ( 15a + 3b = 14 ) ? 3 = 5a + B = 4Y If I use the equation above 6a + B = 6. I can take my latest equation and deduct it from it to happen ‘a’ . So, _6a + B = 6 5a + B = 4Y a = 15 Now that we nave obtained ‘a’ we can now replace its value into the equation to happen the other values. We can now happen ‘b’ by replacing in 15 into the equation as follows: 5 ( _ ) + B = 4Y 5 ( _ ) = 6Y B = 4Y – 6Y b= -2 We have now found values a & b. I will now try to happen values c and d by replacing in the two values I now possess. So we will now stand in. in the values to the equation If we take sequence 2 for our value of N with our present values we will acquire: _ ( 8 ) – 2 ( 4 ) + hundred = 8 this can so be simplified to, _ ( 2 ) – 2 ( 2 ) + hundred = 4 by spliting the equation by two. We will go on with the computation utilizing the simplified version of the equation ; To happen ‘c’ we will utilize the undermentioned computations: 1 ) _ ( 4 ) – 2 ( 2 ) + hundred = 4 2 ) 55 – 4 + c = 4 3 ) 55 + degree Celsiuss = 8 4 ) degree Celsius = 8 – 55 5 ) degree Celsius = 2Y Therefore my four values are: a = 15 B = -2 c = 2Y vitamin D = -1 My equation for the working out of the figure of regular hexahedrons in a 3D version of the form is: _ ( n? ) – 2n²+ 2Y ( N ) – 1 Testing out the New Equation I will take sequence 10 to seek and prove this new equation. Sequence 5: N = 5 _ ( 5? ) – 2 ( 5²) + 2Y ( 5 ) – 1 = 166Y – 50 + 135 – 1 = 129 The expression on this sequence seems to be successful. I will now use it to another sequence to be 100 % correct: N = 4 _ ( 4? ) – 2 ( 4²) + 2Y ( 4 ) – 1 = 855 – 32 + 10Y – 1 = 63 The expression once more proves to be successful. Using the expression to happen the figure of squares in a higher sequuence non yet explored in this probe. Sequence 10: N = 10 _ ( 10? ) – 2 ( 10²) + 2Y ( 10 ) – 1 = 13335 – 200 + 26Y – 1 = 1159 Sequence 15: N = 15 _ ( 15? ) – 2 ( 15²) + 2Y ( 15 ) – 1 = 4500 – 450 + 40 – 1 = 4089 This equation has right given me the figure of squares in each sequence which once more proves it can be applied to any of the 3D sequences to give the correct reply. My Conclusions I have made a figure of decisions from the probe I have carried out. First I have deciferred that the equation used in the 2D form was a quadratic. This can be proven through the fact that the 2nd difference was a changeless, a necessary component of any quadratic and besides the fact that the first value has to be squared. This can besides be proved by exemplifying the equation on the graph, making a curve. I have besides established that the top triangular half of the 2D form ever turns out to be a square figure. If we now look at the 3D form, the equation I achieved for it has turned out to a three-dimensional equation. This can be proven through the changeless, once more a necessary feature of any three-dimensional equation and besides the fact that its 1st value must be cubed and its 2nd squared. If we drew a graph we would acquire a ccurved graph in which the line falls steeply, degrees off and so falls once more. The Differentiation Method developed by Jean Holderness played a really of import function in this probe. It helped us to derive cognition of any form and anything that would assist in the invetigation, giving us our changeless, but most significantly it gave us the equation on which to establish our solutions. It was: an²+ bn + c This proven really helpful. To happen our equation we so substituted in different values which we could happen in our distinction tabular array. I have concluded that both the equations proved to be really successful. Therefore the equations are: For the 2D form the equation is ; 2n²– 2n + 1 For the 3D form the equation is ; _ ( n? ) – 2n² + 2Y(n) – 1