GRE Subject Test: Math : Groups

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

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Example Questions

Example Question #1 : Groups

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

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

Possible Answers:

\(\displaystyle \frac{1}{6!}\)

\(\displaystyle \frac{1}{7!}\)

\(\displaystyle 1\)

\(\displaystyle \frac{1}{5!}\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle \frac{1}{7!}\)

Explanation:

The fifth degree Taylor polynomial approximating \(\displaystyle sin(x)\) centered at \(\displaystyle x=0\) is: 

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

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

We need only evaluate this at \(\displaystyle x=1\) and thus we obtain \(\displaystyle \frac{1^7}{7!} = \frac{1}{7!}\)

Example Question #2 : Groups

Which of the following series does not converge?

Possible Answers:

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

\(\displaystyle \sum_{i=0}^{\infty} 0.9999999999999^n\)

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

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

\(\displaystyle \sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}\)

Correct answer:

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

Explanation:

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

 

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

 

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

 

\(\displaystyle n^2\) will dominate over \(\displaystyle (n+1)\) 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 \(\displaystyle n!\) is much greater than \(\displaystyle n^2\), and thus having \(\displaystyle n!\) in the numerator will make the series diverge by the \(\displaystyle n^{th}\) 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.

 

 

 

Example Question #1 : Lagrange Multipliers

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

Possible Answers:

\(\displaystyle f(4.46,11.14)\) is a maximum

\(\displaystyle f(-4.46,-11.14)\) is a minimum

There are no maximums or minimums

\(\displaystyle f(-4.46,11.14)\) is a maximum

\(\displaystyle f(4.46,-11.14)\) is a minimum

 

\(\displaystyle f(4.46,-11.14)\) is a maximum

\(\displaystyle f(-4.46,11.14)\) is a minimum

\(\displaystyle f(4.46,11.14)\) is a maximum

\(\displaystyle f(4.46,-11.14)\) is a minimum

Correct answer:

\(\displaystyle f(4.46,-11.14)\) is a maximum

\(\displaystyle f(-4.46,11.14)\) is a minimum

Explanation:

First we need to set up our system of equations.

\(\displaystyle 2=2\lambda x \rightarrow x=\frac{1}{\lambda}\)

\(\displaystyle -5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}\)

\(\displaystyle x^2+y^2=144\)

Now lets plug in these constraints.

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

 \(\displaystyle \frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144\)

\(\displaystyle \frac{29}{4\lambda^2}=144\)

Now we solve for \(\displaystyle \lambda\)

\(\displaystyle \frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}\)

If

 \(\displaystyle \lambda=\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46\)\(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14\)

 

If

 \(\displaystyle \lambda=-\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46\)\(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14\)

 

Now lets plug in these values of \(\displaystyle x\), and \(\displaystyle y\) into the original equation.

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

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

 

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

Example Question #1 : Applications Of Partial Derivatives

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

Possible Answers:

\(\displaystyle 0\)

\(\displaystyle 1\)

\(\displaystyle 2\sqrt2\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \sqrt2\)

Correct answer:

\(\displaystyle \frac{1}{2}\)

Explanation:

Let \(\displaystyle g = x^2 +2y^2\)To find the absolute minimum value, we must solve the system of equations given by

\(\displaystyle \triangledown f = \lambda\triangledown g, g =1\).

So this system of equations is

\(\displaystyle f_x = \lambda g_x\), \(\displaystyle f_y = \lambda g_y\), \(\displaystyle g =1\).

Taking partial derivatives and substituting as indicated, this becomes

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

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

On the other hand, if instead \(\displaystyle \lambda =1\), this forces \(\displaystyle y = 0\) from the 2nd equation, and \(\displaystyle x = \pm 1\) from the 3rd equation. This gives us two more extreme candidate points; \(\displaystyle (-1,0),(1,0)\).

 

Taking all four of our found points, and plugging them back into \(\displaystyle f\), we have

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

Hence the absolute minimum value is \(\displaystyle \frac{1}{2}\).

 

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