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GRE Subject Test: Math : Groups

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Example Questions

Example Question #1 : Abstract Algebra

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for f(x) = sin(x) , centered at x = 0 .

What is the Lagrange error of the polynomial approximation to sin(1) ?

Possible Answers:

$\frac{1}{7!}$

$\frac{1}{5!}$

$\frac{1}{6!}$

Correct answer:

$\frac{1}{7!}$

Explanation:

The fifth degree Taylor polynomial approximating sin(x) centered at x=0 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 $\frac{1^7}{7!} = \frac{1}{7!}$

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Example Question #2 : Abstract Algebra

Which of the following series does not converge?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

n^2 will dominate over (n+1) 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 n^2 , 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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Example Question #1 : Applications Of Partial Derivatives

Find the minimum and maximum of f(x,y)=2x-5y , subject to the constraint x^2+y^2=144 .

Possible Answers:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

f(4.46,11.14) is a maximum

f(-4.46,-11.14) is a minimum

There are no maximums or minimums

f(-4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

Correct answer:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Explanation:

First we need to set up our system of equations.

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

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

x^2+y^2=144

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.

f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62

f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62

We can conclude from this that f(4.46,-11.14) is a maximum, and f(-4.46,11.14) is a minimum.

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Example Question #1 : Lagrange Multipliers

Find the absolute minimum value of the function f(x,y) = x^2+y^2 subject to the constraint x^2 +2y^2 = 1 .

Possible Answers:

$\frac{1}{2}$

$\sqrt2$

$2\sqrt2$

Correct answer:

$\frac{1}{2}$

Explanation:

Let g = x^2 +2y^2 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

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , g =1 .

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 x=0 or $\lambda = 1$ . If x=0 , 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 y = 0 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; (-1,0),(1,0) .

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 $\frac{1}{2}$ .

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GRE Subject Test: Math : Groups

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Abstract Algebra

Example Question #2 : Abstract Algebra

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Lagrange Multipliers

All GRE Subject Test: Math Resources

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