MathTank - User contributions [en-ca]

Area bounded by circle

2026-04-19T18:23:15Z

Alexandk:

Suppose we take for granted (as we often do) that the circumference of a circle of radius <math>r</math> is <math>2\pi r</math>. Can we use this fact to show that the area bounded by a circle is <math>\pi r^2</math>

Area bounded by circle

2026-04-19T15:25:39Z

Alexandk:

Suppose we take for granted (as we often do) that the circumference of a circle of radius [math]r[/math] is $2\pi r$. Can we use this fact to show that the area bounded by a circle is $\pi r^2$?

Area bounded by circle

2026-04-19T15:23:31Z

Alexandk: page Area bounded by circle was created

Suppose we take for granted (as we often do) that the circumference of a circle of radius $r$ is $2\pi r$. Can we use this fact to show that the area bounded by a circle is $\pi r^2$?

Single-Variable Calculus

2026-04-19T15:20:34Z

Alexandk: Link to area bounded by circle added

[[Area bounded by circle]]

[[Derivative rules via approximations|Derivative rules using approximation]]

[[Domino effect and tractrix]]

[[test|Introduction to limits]]

[[Number e and compound interest]]

[[Piece-wise defined functions]]

[[Stacking with harmonic series]]

Main Page

2022-11-09T00:13:31Z

Alexandk:

Welcome to MathTank!

MathTank is a collection of resources for instructors who teach mathematics and other STEM-related subjects at an undergraduate level. Our goal is to provide you with notes and examples that will motivate and engage your students. [[Derivative rules via approximations|Here]] is an example.

MathTank is created by [https://uts.nipissingu.ca/alexandk/ Alexandre Karassev,] [https://www.nipissingu.ca/academics/faculty-arts-and-science/mathematics Mathematics] Professor at [https://www.nipissingu.ca/ Nipissing University.] Currently, only a [[Contributors|few selected participants]] can contribute the content and edit pages. However, everybody is welcome to use materials of the site at their discretion. You can start by looking at the list of [[topics]].

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==Getting started==
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Domino effect and tractrix

2022-05-16T20:28:21Z

Alexandk:

Consider a set of dominos standing along a straight line with equal distances between them.
[[File:Domino effect.png|thumb|alt=|none]]
Push the leftmost domino and assume they all fall except for the very last one. What curve do their tops trace?
[[File:Domino effect2.png|none|thumb]]
To answer this question, imagine that the dominos are infinitely thin and thus replace each domino with a stick (of the same length). Then it is reasonable to assume that these sticks are tangent to the "ideal" curve that follows the tops of the dominos.
[[File:Domino effect tangents.png|none|thumb]]
Thus, we need to find an equation of the cure that has the following property: segments of the tangent line, "trapped" between the curve and the ''x''-axis, have constant length, call it ''a''.

File:Domino effect tangents.png

2022-05-16T20:26:34Z

Alexandk:

Dominos, tangents

Domino effect and tractrix

2022-05-16T20:25:39Z

Alexandk:

File:Domino effect2.png

2022-05-16T20:18:40Z

Alexandk:

Domino effect

Domino effect and tractrix

2022-05-16T20:18:07Z

Alexandk:

Domino effect and tractrix

2022-05-16T20:17:44Z

Alexandk:

Consider a set of dominos standing along a straight line with equal distances between them.
[[File:Domino effect.png|left|thumb]]
Push the leftmost domino and assume they all fall except for the very last one. What curve do their tops trace?

Domino effect and tractrix

2022-05-16T20:16:52Z

Alexandk:

Consider a set of dominos standing along a straight line with equal distances between them.
[[File:Domino effect.png|left|thumb]]

File:Domino effect.png

2022-05-16T20:16:24Z

Alexandk:

Domino, standing

Domino effect and tractrix

2022-04-15T14:13:09Z

Alexandk: Created page with "Consider a set of dominos standing along a straight line with equal distances between them."

Consider a set of dominos standing along a straight line with equal distances between them.

Single-Variable Calculus

2022-04-15T13:46:08Z

Alexandk:

[[Derivative rules via approximations|Derivative rules using approximation]]

[[Domino effect and tractrix]]

[[test|Introduction to limits]]

[[Number e and compound interest]]

[[Piece-wise defined functions]]

[[Stacking with harmonic series]]

Piece-wise defined functions

2022-04-04T01:15:03Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

a) Assuming conservation of energy and using symmetry, we see that

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1)\sqrt{3/g})^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height <math> h.</math> Namely we get <math> x = 3t </math> and <math> y = h - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math> Solving this in terms of <math> x,</math> we obtain <math> y = h - gx^2/18,</math> <math> 0\le x \le 3\sqrt{2h/g}, </math> and

<math> y =
\begin{cases}
h - gx^2/18, & 0\le x \le 3\sqrt{2h/g}\\
h - g(x-6(k+1)\sqrt{2h/g})^2/18, & (3+6k)\sqrt{2h/g} \le x \le (9+6k)\sqrt{2h/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

Piece-wise defined functions

2022-04-04T01:08:45Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

a) Assuming conservation of energy and using symmetry, we see that

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1)\sqrt{3/g})^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height <math> h.</math> Namely we get <math> x = 3t </math> and <math> y = h - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math> Solving this in terms of <math> x,</math> we obtain <math> y = h - gx^2/18,</math> <math> 0\le x \le 3\sqrt{2h/g}, </math> and

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1)\sqrt{2h/g})^2/18, & (3+6k)\sqrt{2h/g} \le x \le (9+6k)\sqrt{2h/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

b) If 20% of energy is lost on each bounce, then the maximal height the ball reaches after the first bounce is 80% of the initial height, i.e. <math> 1.5\cdot 0.8. </math> Similarly, the height after the <math> k+ 1 </math> bounce will be <math> 1.5\cdot(0.8)^{k+1},</math> <math> k=0,1,2,\dots </math> Thus, for <math> 3\sqrt{3/g} \le x \le 9\sqrt{3/g} </math> we have <math> y = 1.5 - g(x-6)^2/18 </math>

Piece-wise defined functions

2022-04-04T01:05:36Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

a) Assuming conservation of energy and using symmetry, we see that

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1)\sqrt{3/g})^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height <math> h.</math>

b) If 20% of energy is lost on each bounce, then the maximal height the ball reaches after the first bounce is 80% of the initial height, i.e. <math> 1.5\cdot 0.8. </math> Similarly, the height after the <math> k+ 1 </math> bounce will be <math> 1.5\cdot(0.8)^{k+1},</math> <math> k=0,1,2,\dots </math> Thus, for <math> 3\sqrt{3/g} \le x \le 9\sqrt{3/g} </math> we have <math> y = 1.5 - g(x-6)^2/18 </math>

Piece-wise defined functions

2022-04-04T01:04:00Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

a) Assuming conservation of energy and using symmetry, we see that

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1)\sqrt{3/g})^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

b) If 20% of energy is lost on each bounce, then the maximal height the ball reaches after the first bounce is 80% of the initial height, i.e. <math> 1.5\cdot 0.8. </math> Similarly, the height after the <math> k+ 1 </math> bounce will be <math> 1.5\cdot(0.8)^{k+1},</math> <math> k=0,1,2,\dots </math> Thus, for <math> 3\sqrt{3/g} \le x \le 9\sqrt{3/g} </math> we have <math> y = 1.5 - g(x-6)^2/18 </math>

Piece-wise defined functions

2022-04-04T00:57:03Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

a) Assuming conservation of energy and using symmetry, we see that

<math> y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - g(x-6(k+1))^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
\end{cases}
</math>

Piece-wise defined functions

2022-04-04T00:50:50Z

Alexandk:

Piece-wise defined functions

2022-04-04T00:45:52Z

Alexandk:

A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example...
(examples may vary).

Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).

''' Pool table '''

A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.

'''Even numbers'''

Determine how many positive even numbers are there thad do not exceed a given positive integer.

'''Bungee jumping'''

Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).

Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when <math> 0\le x \le 10 </math>, the only acting force is gravity, so the speed is <math> gt </math>, or, equivalently, <math> v(x) = \sqrt{2gx} </math>. After the point where <math> x = 10 </math>, a restoring force comes into play. Velocity at that point is <math> v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} </math>. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when <math> x = 10+ mg/k.</math> If <math> \triangle x </math> denotes the displacement from the equilibrium, the velocity is <math> v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} </math>, where <math> A </math> is the amplitude and can be computed by substituting <math> \triangle x = - mg/k </math> and therefore (using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, and <math> g\approx 9.8</math>m/s2) is
<math> A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5.</math>

Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get

<math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math>

for <math> 10 \le x \le 10+(mg/k)+A.</math>

Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s2, and <math> A \approx 4.5,</math> we obtain

<math> v(x) =
\begin{cases}
\sqrt{2gx}, & 0\le x \le 10 \\
\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
\end{cases}
</math>

''' Ball trajectory'''

A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 3 - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math>

Number e and compound interest

2022-04-03T20:03:55Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^{12} </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:

<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n.</math>

Number e and compound interest

2022-04-03T20:03:15Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:

<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n.</math>

Number e and compound interest

2022-04-03T20:02:00Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:

<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

Number e and compound interest

2022-04-03T20:00:53Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>)

Number e and compound interest

2022-04-03T19:59:55Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously". It can be shown that the above limit exists and is approximately equal to <math> 2.7.</math>

Number e and compound interest

2022-04-03T19:58:57Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^{12|} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously".

Number e and compound interest

2022-04-03T19:58:11Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> \left(1+\frac{1}{12}\right)^12 \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously".

Number e and compound interest

2022-04-03T19:57:44Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> (1+\frac{1}{12}\right)^12 \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> \left(1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty \left(1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded "continuously".

Number e and compound interest

2022-04-03T19:57:14Z

Alexandk:

Suppose you invest some amount <math> P </math> for a period of time (that can be a week, month, year etc.) at effective rate of compound interest <math> i. </math> At the end of one such period you have <math> P +iP = P(1+i)</math> accumulated in your account. If you keep your investment for two such periods, the value in your account will be <math> P(1+i) + i[P(1+i)] = P(1+i)^2. </math> In general, if the investment is for <math> n </math> periods, at the end your account will have accumulated <math> P(1+i)^n. </math>

Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have

<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>

in your account. Using the Binomial formula

<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>

we see that

<math> (1+\frac{1}{12}\right)^12 \ge 1 + 12\cdot \frac{1}{12} = 2 </math>

and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.

Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be

<math> (1+\frac{1}{n}\right)^n. </math>

For very large values of <math> n </math> we therefore can consider the limit

<math> \lim _{n\to\infty} (1+\frac{1}{n}\right)^n </math>

which corresponds to the interest compounded `` continuously".

Number e and compound interest

2022-04-03T19:52:38Z

Alexandk:

Number e and compound interest

2022-04-03T19:52:16Z

Alexandk:

Number e and compound interest

2022-04-03T19:50:31Z

Alexandk:

Number e and compound interest

2022-04-03T19:43:03Z

Alexandk:

Number e and compound interest

2022-04-03T19:35:14Z

Alexandk: Created page with "Suppose you invest some amount for a period of time at effective rate of interest"

Suppose you invest some amount for a period of time at effective rate of interest

Single-Variable Calculus

2022-04-03T19:32:22Z

Alexandk:

[[Derivative rules via approximations|Derivative rules using approximation]]

[[test|Introduction to limits]]

[[Number e and compound interest]]

[[Piece-wise defined functions]]

[[Stacking with harmonic series]]

Piece-wise defined functions

2022-04-02T18:03:15Z

Alexandk:

Piece-wise defined functions

2022-04-02T18:01:08Z

Alexandk:

Piece-wise defined functions

2022-04-02T18:00:46Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:38:03Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:36:49Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:36:21Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:34:48Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:18:50Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:17:37Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:17:15Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:15:15Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:12:14Z

Alexandk:

Piece-wise defined functions

2022-04-02T16:11:14Z

Alexandk: