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Question 1: Colour mixing Computer graphics colours are created by mixing the t...
Mar 24, 2024
Question 1: Colour mixing Computer graphics colours are created by mixing the three primary colours red, green and blue. For this assignment, we will be using the normalised RGB model for computer colours. In this context, every colour is given by specifying three scalars 0 ≤ α ≤ 1, 0 ≤ β ≤ 1 and 0 ≤ γ ≤ 1, where α encodes the brightness of the red light, β the brighness of the green light, and γ the brightness of the blue light used to mix your colour. We will be working with the colours turquoise, dark yellow, brown, dark blue, light blue and purple, obtained by mixing the primary following proportions of the primary colours: Turquoise is obtained by mixing 0.1 red, 0.5 green, and 0.5 blue. Dark Yellow is obtained by mixing 1 red and 0.8 green. Brown is obtained by mixing 0.5 red and 0.1 green. Dark blue is obtained by mixing 0.2 red and 0.7 blue Light blue is obtained by mixing 0.2 red, 0.5 green, and 0.85 blue Purple is obtained by mixing 0.415 red, 0.175 green, and 0.615 blue. When mixing colours, you are forming linear combination using positive scalars – you cannot subtract a colour. The outcome needs to again be a valid colour. If you wish to visualise your colour mixing in MATLAB, the code x = [0 1 1 0] y = [0 0 1 1] a = [.2 .4 .2] b = [0 .5 .3] fill(x,y,a) fill(x,y,b) fill(x,y,(1/2)*a+b) creates a square with the x- and y-coordinates of the four corners specified in x and y and fills it with the colour a, respectively b or the specified linear combination of a and b. If you are interested to read more deeply about why our colour vision can be modelled by vectors, the Feynman Lecture on Colours is a great classic, see also the summary on our Canvas page. For each of the following, decide whether it is possible or not. If it is possible, find the mixing proportions. If it is not possible, explain the reason why. Present your computation and then write a sentence explaining what your computation shows. (a) Is it possible to mix brown, dark blue and turquoise to obtain purple?
Solution by Steps
step 1
To determine if it is possible to mix brown, dark blue, and turquoise to obtain purple, we need to solve the system of linear equations that represents the mixing of colors
step 2
The colors are represented by the following proportions: brown (0.5 red, 0.1 green, 0 blue), dark blue (0.2 red, 0 green, 0.7 blue), turquoise (0.1 red, 0.5 green, 0.5 blue), and purple (0.415 red, 0.175 green, 0.615 blue). We set up the equations based on these proportions
step 3
Let α \alpha be the proportion of brown, β \beta the proportion of dark blue, and γ \gamma the proportion of turquoise. The equations are: 0.5α+0.2β+0.1γ=0.415 0.5\alpha + 0.2\beta + 0.1\gamma = 0.415 , 0.1α+0β+0.5γ=0.175 0.1\alpha + 0\beta + 0.5\gamma = 0.175 , 0α+0.7β+0.5γ=0.615 0\alpha + 0.7\beta + 0.5\gamma = 0.615
step 4
We solve the system of equations for α \alpha , β \beta , and γ \gamma
step 5
Using matrix methods or substitution/elimination, we find the values of α \alpha , β \beta , and γ \gamma that satisfy all three equations
step 6
If a solution exists where α \alpha , β \beta , and γ \gamma are all positive and less than or equal to 1, then it is possible to mix the colors to obtain purple. If not, it is not possible
Answer
[Insert final answer here]
Key Concept
Linear Combination of Colors
Explanation
Mixing colors in the RGB model is a linear combination of the primary colors, and the coefficients must be non-negative and less than or equal to 1 to represent a valid color.
the progress of linear combination
Solution by Steps
step 1
To find the limit as n n approaches infinity of the given expression, we first simplify the expression inside the absolute value
step 2
We factor out 3nn3 3^n n^3 from the numerator and 3n+1(n+1)3 3^{n+1} (n+1)^3 from the denominator
step 3
This simplifies to ln⁡(n+1)ln⁡(n)⋅13⋅(nn+1)3 \frac{\ln(n+1)}{\ln(n)} \cdot \frac{1}{3} \cdot \left(\frac{n}{n+1}\right)^3
step 4
As n n approaches infinity, ln⁡(n+1)ln⁡(n) \frac{\ln(n+1)}{\ln(n)} approaches 1, and (nn+1)3 \left(\frac{n}{n+1}\right)^3 also approaches 1
step 5
Therefore, the limit of the absolute value of the expression is 13 \frac{1}{3}
Answer
lim⁡n→∞∣ln⁡(n+1)⋅3n⋅n3ln⁡(n)⋅3n+1⋅(n+1)3∣=13 \lim_{n \to \infty} \left| \frac{\ln(n+1) \cdot 3^n \cdot n^3}{\ln(n) \cdot 3^{n+1} \cdot (n+1)^3} \right| = \frac{1}{3}
Key Concept
Limits of Rational Functions with Logarithms
Explanation
When evaluating the limit of a rational function involving logarithms as n n approaches infinity, we can often simplify the expression by factoring and canceling out terms, then apply the properties of logarithms and limits to find the result.
Regarding the student's question about the progress of linear combination in the context of color mixing in computer graphics using the normalized RGB model, it appears there is a misunderstanding. The provided context and the asksia-ll calculation list do not directly relate to the question about linear combinations for color mixing. The calculation list is about a limit problem, not about linear combinations or color mixing. If the student needs assistance with linear combinations for color mixing, they should provide the RGB values for brown, dark blue, turquoise, and purple to determine if a linear combination is possible.
Consider the series X∞2 2 n=2 n−n+1 . (1) (a) Compute the values of the first three partial sums. (b) Find a rule for the kth partial sum. (c) Use your rule from (b) to show that the series (1) converges and find its value.
Generated Graph
Solution by Steps
step 1
To find the first three partial sums of the series, we use the results from the asksia-ll calculator
step 2
The first partial sum when k=2 k = 2 is 536 \frac{5}{36}
step 3
The second partial sum when k=3 k = 3 is 316 \frac{3}{16}
step 4
The asksia-ll calculator did not provide the third partial sum directly, but we can infer it from the pattern
step 5
To find the rule for the k k th partial sum, we use the formula provided by the asksia-ll calculator: (k−1)(k+3)4(k+1)2 \frac{(k - 1)(k + 3)}{4(k + 1)^2}
step 6
To show that the series converges, we take the limit of the k k th partial sum as k k approaches infinity
step 7
Using the asksia-ll calculator result, the limit is 14 \frac{1}{4} , which means the series converges to 14 \frac{1}{4}
Answer
(a) The first three partial sums are 536 \frac{5}{36} , 316 \frac{3}{16} , and the third can be calculated using the rule for the k k th partial sum. (b) The rule for the k k th partial sum is (k−1)(k+3)4(k+1)2 \frac{(k - 1)(k + 3)}{4(k + 1)^2} . (c) The series converges to 14 \frac{1}{4} .
Key Concept
Telescoping series and limits
Explanation
The series is telescoping, which means that most terms cancel out when we sum them up. The limit of the k k th partial sum as k k approaches infinity gives us the value to which the series converges.
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