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Tutor profile: Parth B.

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Parth B.
Business Intelligence Application Developer at Amtex
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Questions

Subject: SAT II Mathematics Level 2

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Question:

What is the smallest distance between the point(-2,-2) and a point on the circumference of the circle given by (x - 1)^2 + (y - 2)^2 = 4?

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Parth B.
Answer:

The equation of a circle is given by (x−h)^2+(y−k)^2=r^2 where (h,k) is the centre of the circle and r is the radius of the circle (r≥0). Your equation (x−1)^2+(y−2)^2=4 therefore has a centre of (1,2) and a radius of 2. NOTE : The given point now clearly lies outside the circle. That is, it is in the bottom left quadrant. From here we need to simply find the distance between two points ( the center of the circle and the given point ). From there we can simply subtract from this distance, the radius of our circle to get the required distance. -- Finding distance between center of circle(1,2) and (-2,-2) =√(2−(−2))^2+(1−(−2))^2 =√42+32 =√16+9 =√25 =5 From this we subtract the radius of circle, which is sqrt(4) to get 2. Then 5-2 =3 3 is the final answer. Questions like these can get tricky when you are under the pressure of finishing them under the time limit of the test. A lot of people leave the question after solving for the distance '5' and assume that to be the right answer in a hurry. There are many other such tricky questions we can work through which are easy to score points in, only if we understand the simple trick that goes into solving them.

Subject: Python Programming

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Question:

Let’s say I give you a list saved in a variable: a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]. Write one line of Python that takes this list a and makes a new list that has only the even elements of this list in it.

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Parth B.
Answer:

This exercise focuses on your knowledge of the Python language itself and requires you to be well-versed with the basic usage of loops and lists. The answer is : b = [element for element in a if element % 2 == 0] Python was made to be read as a rather simple programming language. Here we are going through a given list 'a' and rather than taking the long route of writing a full fledged if-condition we are using the "shorter" form available to us in the language. This way of writing for loops has now been adopted into other modern languages as well. Read the code as following: 1) Take list 'a' 2) For each 'element' in 'a' ( think let element = a[0] or a[1] ). The 'for' statement will do this for all elements in 'a' 3) Take the element only if element%2==0 ( That is only take if the element is divisible by two, that is, it is an even number ) 4) Store element in list b ( this is why we write element in the beginning of the code ) A complete solution would look like this : a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] b = [number for number in a if number % 2 == 0] print(b)

Subject: Calculus

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Question:

For the following Power series, determine the interval and radius of convergence $$ \sum\limits_{n = 0}^\infty {\frac{{n + 1}}{{\left( {2n + 1} \right)!}}{{\left( {x - 2} \right)}^n}} $$

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Parth B.
Answer:

To begin, we first need to do the "Ratio Test" on the given equation. The Ratio test requires that we add 1 to the variable that we are using for the limit. Here that variable is 'n'. This should give us a new equation which we then need to divide by the initial equation. This step will look like this : $$ \begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right){{\left( {x - 2} \right)}^{n + 1}}}}{{\left( {2n + 3} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right){{\left( {x - 2} \right)}^n}}}} \right| \end{align*}$$ The new equation should be simple to derive, we have just taken each 'n' in the equation and replaced it with 'n+1' Note that the initial equation is now in inverse form. This is ok because we are dividing the new equation by the initial equation. -- We can now further simplify this equation, then it will look like this. $$ \begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right){{\left( {x - 2} \right)}^{n + 1}}}}{{\left( {2n + 3} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right){{\left( {x - 2} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right)\left( {x - 2} \right)}}{{\left( {2n + 3} \right)\left( {2n + 2} \right)\left( {2n + 1} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right)}}} \right|\\ \end{align*} $$ This was a fairly simple simplification of the equation where we have divided the powers of (x-2) and worked with the provided factorials. From here on, we can further simplify the equation and it will now look like this: $$ \begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right){{\left( {x - 2} \right)}^{n + 1}}}}{{\left( {2n + 3} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right){{\left( {x - 2} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right)\left( {x - 2} \right)}}{{\left( {2n + 3} \right)\left( {2n + 2} \right)\left( {2n + 1} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right)}}} \right|\\ & = \left| {x - 2} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{n + 2}}{{\left( {2n + 3} \right)\left( {2n + 2} \right)\left( {n + 1} \right)}} = 0\end{align*} $$ Here we can easily see the L=0 for every value of x. This is because the power of 'n' in the denominator is much higher than that of the numerator. This means that when 'n' will tend to infinity, the equation will come out to be zero. Therefore, we know that the interval of convergence is $$ \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty < x < \infty }} $$ and radius of convergence is $$ \require{bbox} \bbox[2pt,border:1px solid black]{{R = \infty }} $$

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