# GRE Subject Test: Math : Groups

## Example Questions

### Example Question #1 : Lagrange Error

Let  be the fifth-degree Taylor polynomial approximation for , centered at .

What is the Lagrange error of the polynomial approximation to ?

Possible Answers:

Correct answer:

Explanation:

The fifth degree Taylor polynomial approximating  centered at  is:

The Lagrange error is the absolute value of the next term in the sequence, which is equal to .

We need only evaluate this at  and thus we obtain

### Example Question #1 : Alternating Series With Error Bound

Which of the following series does not converge?

Possible Answers:

Correct answer:

Explanation:

We can show that the series   diverges using the ratio test.

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  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.

### Example Question #1 : Applications Of Partial Derivatives

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

Possible Answers:

is a maximum

is a minimum

is a maximum

is a minimum

is a maximum

is a minimum

is a maximum

is a minimum

There are no maximums or minimums

Correct answer:

is a maximum

is a minimum

Explanation:

First we need to set up our system of equations.

Now lets plug in these constraints.

Now we solve for

If

If

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.

### Example Question #1 : Lagrange Multipliers

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

Possible Answers:

Correct answer:

Explanation:

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

.

So this system of equations is

, , .

Taking partial derivatives and substituting as indicated, this becomes

.

From the left equation, we see either or . If , then substituting this into the other equations, we can solve for , and get , , giving two extreme candidate points at .

On the other hand, if instead , this forces from the 2nd equation, and 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

.

Hence the absolute minimum value is .