Showing posts with label calculus. Show all posts
Showing posts with label calculus. Show all posts
Saturday, April 21, 2018
AM-GM-HM
From OSP SMA 2018
For positive $a,b,c$ such that $1/a + 1/b + 1/c = 3$ prove that:
$$a+b+c + \frac{4}{1+(abc)^\frac{2}{3}} \geq 5$$
Labels:
Algebra,
AM-GM,
calculus,
factorization,
Inequality,
polynomial
Friday, March 23, 2018
Integral and Inequality
Suppose $f(x)$ is a continuously differentiable function on $[a,b]$ satisfying:
$$f(a) = f(b) = 0$$
$$\int_a^b (f(x))^2 dx = 1$$
Then show that:
$$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$
Solution
By Cauchy:
$$\int_a^b x^2(f'(x))^2 dx = \int_a^b (f(x))^2 dx . \int_a^b x^2(f'(x))^2 dx \geq (\int_a^b xf'(x)f(x)dx)^2$$
The last integral can be evaluated using integration by parts, using $u=x$ and $dv = f'(x)f(x)dx$ which yields $v = (f(x))^2/2$, so that:
$$\int_a^b xf'(x)f(x)dx = \frac{bf^2(b) - af^2(a)}{2} - \frac{1}{2} \int_a^b (f(x))^2 dx = -\frac{1}{2}$$
so:
$$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$
Monday, December 19, 2011
Having fun with infinite series
1. Warm-up problem: show that
$$1 + \frac{1}{2} + \frac{1}{3} + \cdots = \infty$$
2. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$
3. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2 \log n$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$
$$1 + \frac{1}{2} + \frac{1}{3} + \cdots = \infty$$
2. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$
3. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2 \log n$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$
Solution
Problem 1
This is a standard textbook proof of the divergence of harmonic series, but the point here is to prepare the reader for the subsequent proofs $$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ $$\frac{1}{5} + \dots + \frac{1}{8} > \frac{1}{8} + \dots + \frac{1}{8} = \frac{1}{2}$$ and so on. So the original series clearly diverges to infinity. The main crux of the proof here is this assertion: $$\frac{1}{2^n+1} + \dots + \frac{1}{2^{n+1}} > \frac{1}{2^{n+1}} + \dots + \frac{1}{2^{n+1}} = \frac{1}{2}$$ for each $n$.Problem 2
Similar to the proof above, for each $n$ we have: $$a_{2^n+1} + \dots + a_{2^{n+1}} < 4^{n+1}$$ So by AM-HM we have: $$\frac{1}{a_{2^n+1}} + \dots + \frac{1}{a_{2^{n+1}}} > \frac{4^n}{a_{2^n+1} + \dots + a_{2^{n+1}}} > \frac{1}{4}$$ So the original series is greater than $1/4 + 1/4 + \dots = \infty$Problem 3
Similar to the proof above, for each $n$ we have: $$a_{2^n+1} + \dots + a_{2^{n+1}} < 4^{n+1} \log (2^n) = n . 4^{n+1}.\log 2$$ So by AM-HM we have: $$\frac{1}{a_{2^n+1}} + \dots + \frac{1}{a_{2^{n+1}}} > \frac{4^n}{a_{2^n+1} + \dots + a_{2^{n+1}}} > \frac{1}{4 \log 2 n}$$ So the original series is greater than $\frac{1}{4 \log2} (1 + \frac{1}{2} + \frac{1}{3} + \dots)$ which is also divergent.
Labels:
Algebra,
calculus,
cauchy,
harmonic,
infinite series,
infinite sum,
integral,
sequence
Wednesday, September 1, 2010
Integral Inequality
Prove that:
$$\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}} $$
Solution
Integrate by parts:
$$\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx$$
So
$$\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx$$
And
$$| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|$$
$$\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx$$
$$\leq \int_\pi^\infty \frac{1}{x^2} dx $$
$$= \frac{1}{\pi}$$
$$\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}} $$
Solution
Integrate by parts:
$$\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx$$
So
$$\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx$$
And
$$| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|$$
$$\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx$$
$$\leq \int_\pi^\infty \frac{1}{x^2} dx $$
$$= \frac{1}{\pi}$$
Labels:
calculus,
Inequality,
integral,
integration by parts
Friday, March 12, 2010
Osculating Circle, Ellipse, and Cone
An osculating circle of a point on a curve is defined as a circle that:
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point
In other words, it is a second-degree approximation circle of the curve at that point.
http://en.wikipedia.org/wiki/Osculating_circle
Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.
If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.
Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.
Solution:
http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html
Second Solution:
http://dharmath.blogspot.com/2010/03/second-solution-osculating-circle.html
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point
In other words, it is a second-degree approximation circle of the curve at that point.
http://en.wikipedia.org/wiki/Osculating_circle
Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.
If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.
Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.
Solution:
http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html
Second Solution:
http://dharmath.blogspot.com/2010/03/second-solution-osculating-circle.html
Labels:
3D,
calculus,
cone,
cross section,
ellipse,
Geometry,
osculating circle,
solid geometry,
Solved
Wednesday, January 13, 2010
Infinite fractional sum
Find the limit to the infinite sum:
$$ 1 - \frac{1}{2} +\frac{1}{3} - \frac{1}{4} + \cdots$$
$$ 1 - \frac{1}{2} +\frac{1}{3} - \frac{1}{4} + \cdots$$
Labels:
Algebra,
calculus,
infinite series,
infinite sum,
limit
Monday, December 21, 2009
Average Winning Roll
$n$ people are rolling a random real number between zero and one. The highest roll wins. What is the expected value of the winning roll?
Labels:
average,
calculus,
Combinatorics,
expected value,
Random,
roll
Tuesday, December 15, 2009
Twist And Spin
A matrix $M$ is skew-symmetric if and only if $M^{tr} = -M$. Let $S$ be the set of all 4x4 skew-symmetric matrices.
Suppose $F:\mathbb{R}^4 \to \mathbb{R}^4$ is a differentiable vector field in $\mathbb{R}^4$, we define the operation "twist" that maps each point in $\mathbb{R}^4$ to $S$ as follows:
If $F(x,y,z,w) = A\vec i + B \vec j + C \vec k + D \vec l$ then
$(\bold{twist} F)(x,y,z,w) = \left( \begin{array}{cccc}
0 & B_x - A_y & C_x - A_z & D_x - A_w \\
A_y-B_x & 0 & C_y - B_z & D_y - B_w \\
A_z-C_x & B_z - C_y & 0 & D_z - C_w \\
A_w - D_x & B_w - D_y & C_w - D_z & 0 \end{array} \right) $
Where $B_x$ denotes $\partial B / \partial x$ and so on. One way that makes it easier to memorize the formula is to associate the first column and row with $x$, second column and row with $y$, third with $z$, and last with $w$.
If $G : \mathbb{R}^4 \to S$ is a differentiable function that takes a point in $\mathbb{R}^4$ to a skew-symmetric matrix, we define the operation "spin" that produces a vector field in $\mathbb{R}^4$ as follows:
If $G(x,y,z,w) = \{ \sum_{i,j} G_{ij}(x,y,z,w) \}$ (where $G_{ij}$ is the entry at the $i$-th row and $j$-th column, and since $G$ is skew-symmetric, then $G_{i,i} = 0$ and $G_{ji} = -G_{ij}$)
Then
$(\bold{spin} G)(x,y,z,w) = (G_{23_w} + G_{34_y} + G_{42_z}) \vec i $
$+(G_{13_w} + G_{34_x} + G_{41_z}) \vec j$
$+(G_{12_w} + G_{24_x} + G_{41_y}) \vec k$
$+(G_{12_z} + G_{23_x} + G_{31_y}) \vec l$
where again $G_{ij_x}$ denotes $\partial G_{ij} / \partial x$ and so on.
Prove the following:
1. $(\bold{spin}(\bold{twist} F)) = \vec 0$
2. $\bigtriangledown \cdot ( \bold{spin} G) = 0$
3. If $F$ is a conservative vector field, then $\bold{twist} F = \vec 0$
Twist
Suppose $F:\mathbb{R}^4 \to \mathbb{R}^4$ is a differentiable vector field in $\mathbb{R}^4$, we define the operation "twist" that maps each point in $\mathbb{R}^4$ to $S$ as follows:
If $F(x,y,z,w) = A\vec i + B \vec j + C \vec k + D \vec l$ then
$(\bold{twist} F)(x,y,z,w) = \left( \begin{array}{cccc}
0 & B_x - A_y & C_x - A_z & D_x - A_w \\
A_y-B_x & 0 & C_y - B_z & D_y - B_w \\
A_z-C_x & B_z - C_y & 0 & D_z - C_w \\
A_w - D_x & B_w - D_y & C_w - D_z & 0 \end{array} \right) $
Where $B_x$ denotes $\partial B / \partial x$ and so on. One way that makes it easier to memorize the formula is to associate the first column and row with $x$, second column and row with $y$, third with $z$, and last with $w$.
Spin
If $G : \mathbb{R}^4 \to S$ is a differentiable function that takes a point in $\mathbb{R}^4$ to a skew-symmetric matrix, we define the operation "spin" that produces a vector field in $\mathbb{R}^4$ as follows:
If $G(x,y,z,w) = \{ \sum_{i,j} G_{ij}(x,y,z,w) \}$ (where $G_{ij}$ is the entry at the $i$-th row and $j$-th column, and since $G$ is skew-symmetric, then $G_{i,i} = 0$ and $G_{ji} = -G_{ij}$)
Then
$(\bold{spin} G)(x,y,z,w) = (G_{23_w} + G_{34_y} + G_{42_z}) \vec i $
$+(G_{13_w} + G_{34_x} + G_{41_z}) \vec j$
$+(G_{12_w} + G_{24_x} + G_{41_y}) \vec k$
$+(G_{12_z} + G_{23_x} + G_{31_y}) \vec l$
where again $G_{ij_x}$ denotes $\partial G_{ij} / \partial x$ and so on.
Problems:
Prove the following:
1. $(\bold{spin}(\bold{twist} F)) = \vec 0$
2. $\bigtriangledown \cdot ( \bold{spin} G) = 0$
3. If $F$ is a conservative vector field, then $\bold{twist} F = \vec 0$
Sunday, November 8, 2009
Friday, October 23, 2009
2 Variable Recursion
This problem is written as an auxiliary lemma for the solution to this problem.
Suppose $f(m,n)$ is a function that is defined for non-negative integers $m,n$ where:
$f(m,n) = 0$ if $m < n$
$f(m,0) = 1 \forall m$
$\displaystyle f(m,n) = \sum_{k=1}^{n} f(k-1,k-1) f(m-k,n-k) + f(m-1,n) \forall m \geq n$
Prove that $f(m,n) = \frac{m-n+1}{m+1} \binom{m+n}{m}$
Suppose $f(m,n)$ is a function that is defined for non-negative integers $m,n$ where:
$f(m,n) = 0$ if $m < n$
$f(m,0) = 1 \forall m$
$\displaystyle f(m,n) = \sum_{k=1}^{n} f(k-1,k-1) f(m-k,n-k) + f(m-1,n) \forall m \geq n$
Prove that $f(m,n) = \frac{m-n+1}{m+1} \binom{m+n}{m}$
Subscribe to:
Posts (Atom)