GRE Quantitative Reasoning - Groups

Find the minimum and maximum of , subject to the constraint .

is a maximum

is a minimum

is a maximum

is a minimum

There are no maximums or minimums

is a maximum

is a minimum

is a maximum

is a minimum

Explanation

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

We can conclude from this that is a maximum, and is a minimum.

Find the minimum and maximum of , subject to the constraint .

is a maximum

is a minimum

is a maximum

is a minimum

There are no maximums or minimums

is a maximum

is a minimum

is a maximum

is a minimum

Explanation

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

We can conclude from this that is a maximum, and is a minimum.

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for , centered at .

What is the Lagrange error of the polynomial approximation to ?

$\frac{1}{7!}$

$\frac{1}{5!}$

$\frac{1}{6!}$

Explanation

The fifth degree Taylor polynomial approximating centered at is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for , centered at .

What is the Lagrange error of the polynomial approximation to ?

$\frac{1}{7!}$

$\frac{1}{5!}$

$\frac{1}{6!}$

Explanation

The fifth degree Taylor polynomial approximating centered at is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$

Find the absolute minimum value of the function subject to the constraint .

$\frac{1}{2}$

$\sqrt2$

$2\sqrt2$

Explanation

Let To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

From the left equation, we see either or $\lambda = 1$ . If , then substituting this into the other equations, we can solve for $y, \lambda$ , and get $y = \pm \sqrt{2}/2$ , $\lambda = 1/2$ , giving two extreme candidate points at $(0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$ .

On the other hand, if instead $\lambda =1$ , this forces from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

Find the absolute minimum value of the function subject to the constraint .

$\frac{1}{2}$

$\sqrt2$

$2\sqrt2$

Explanation

Let To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

On the other hand, if instead $\lambda =1$ , this forces from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

Which of the following series does not converge?

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

$\sum_{i=0}^{\infty} \frac{(-1)^n}{n}$

$\sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$

$\sum_{i=0}^{\infty} 0.9999999999999^n$

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

Explanation

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the $n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Which of the following series does not converge?

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

$\sum_{i=0}^{\infty} \frac{(-1)^n}{n}$

$\sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$

$\sum_{i=0}^{\infty} 0.9999999999999^n$

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

Explanation

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the $n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

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