## Wednesday, December 9, 2015

### High School Math Solutions – Trigonometry Calculator, Trig Identities

In a previous post, we talked about trig simplification. Trig identities are very similar to this concept. An identity is the equality of two expressions. When given a trig identity, you have to prove that both sides of the equation are equivalent. This requires using some skills from trig simplification and learning some new skills. One big difference between trig identities and trig simplification is that you are able to work both sides of the equation to show both sides are equivalent. Just like trig simplification, there is no exact recipe on how to prove trig identities. Memorizing trig identities will make proving trig identities 100 times easier.

Trig identities to memorize:

I’ll go over some tips to help make proving trig identities a little easier, and then we will go through an example step by step, so you can understand the thought process when proving trig identities.

Tips:

1. Don’t work on both sides of the equation at the same time
2. Start on the more complicated side
3. Try converting everything into cosine and sine
4. Try working on the other side if you get stuck
5. Memorize the identities
6. If you get frustrated, take a break and look at it again with fresh eyes

1. We will start on the more complicated side (right side) and convert everything to sine and cosine.
=\frac{1-\frac{\sin^2(x)}{\cos^2(x)}}{2\frac{\sin(x)}{\cos(x)}}
2. It looks a little messy, so let’s simplify it.
=\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}\cdot\frac{\cos(x)}{2\sin(x)}
=\frac{\cos^2(x)-\sin^2(x)}{2\cos(x)\sin(x)}

3. I’m stumped as to what to do next, so we’ll start working on the left side and convert everything to sine and cosine.
=\frac{\cos(2x)}{\sin(2x)}
4. This looks familiar. Let’s use the double angle identities. We know what identity to use for cos(2x) based on what the right side of the equation looks like.
=\frac{\cos^2(x)-\sin^2(x)}{2\ cos(x)\sin(x)}

5. The left side now matches the right side. We’re done!

Proving trig identities take a lot of practice. Let’s look at two more examples.

Proving trig identities can be very tricky. It is important to not give up, if you are stumped. Take a break and come back to the problem. Have a page of the identities out in front of you, so you can see your options. The only way to get better is to practice proving trig identities. Check out Symbolab’s trigonometry practice for more help on the topic.

Until next time,

Leah

## Thursday, December 3, 2015

### Intermediate Math Solutions – Functions Calculator, Function Composition

Function composition is when you apply one function to the results of another function. When referring to applying g(x) to f(x), the function composition is denoted as (f\:\circ\:g)(x),\:or\:f(g(x)).

How does function composition work?

x is only used as a place holder in the function. We can substitute in other values for x. When given the function composition (f\:\circ\:g)(x), we take whatever g(x) is equal to and input it in for x in f(x). This will give us the function composition. It is important to know that (f\:\circ\:g)(x)\ne(g\:\circ\:f)(x).

Make sure that the domain of the inner, first function is respected in the function composition.

Let’s go through an example step by step to help you better understand how to solve function compositions.

Given f(x)=2x+3 and g(x)=-x^2+5,find g(f(x+3)).
* g(f(x+3)=g(x)\circ(f(x+3))

1. Substitute x+3 for x in f(x)

2. Substitute f(x+3) for x in g(x)
g(2x+9)

3. Evaluate g(2x+9)

Function composition is very simple. My only piece of advice is to double check your work! It is so easy to have a simple calculation error that will cost you a point or two. Try your best to avoid this and double check your work. Practice a few tricky problems and you’ll be all set to move on from this topic.

Until next time,

Leah

## Tuesday, November 10, 2015

### Advanced Math Solutions – Laplace Calculator, Laplace Transform

In previous posts, we talked about the four types of ODE - linear first order, separable, Bernoulli, and exact. In today’s post, we will learn about Laplace Transforms, how to compute Laplace transforms and inverse Laplace transforms. I’ll admit I was afraid of Laplace transforms before I learned it. The new symbols and the messiness of the problem really intimidated me. However, once I learned what Laplace transforms are and how to do these types of problems, I began to realize that it isn’t so scary. With the assistance of a table and some formulas, anyone can do Laplace transforms.

What is a Laplace Transform?

Laplace transforms can be used to solve differential equations. They turn differential equations into algebraic problems.

Definition:
Suppose f(t) is a piecewise continuous function, a function made up of a finite number of continuous pieces. The Laplace transform of f(t) is denoted L{f(t)} and defined as:
L\left\{f(t)\right\}=\int_{0}^{\infty}e^{-st}f(t)dt
If you see L\left\{f(t)\right\}=\int_{-\infty}^{\infty}e^{-st}f(t)dt, then you can assume that for t\lt0,\quad f(t)=0, and then you can use the original definition of Laplace transform.

Now, we will get into how to compute Laplace transforms:

Laplace transforms can be computed using a table and the linearity property, “Given f(t) and g(t) then, L\left\{af(t)+bg(t)\right\}=aF(s)+bG(s).” The statement means that after you’ve taken the transform of the individual functions, then you can add back any constants and add or subtract the results.

Look at the table and see what functions you can transform. Algebraic manipulation may be required. The table will be your savior when it comes to these problems.

This problem is very simple. It requires looking at the different functions, finding the corresponding transforms in the table, and then adding any constants and adding and subtracting the results together.

Let’s get into inverse Laplace transforms!

What is an inverse Laplace transform?

An inverse Laplace transform is when we are given a transform, F(s), and asked what function(s) we had originally.
Definition:
f(t)=L^{-1}\left\{F(s)\right\}

How to compute the inverse Laplace transforms:

Just like Laplace transforms have a linearity property, so do inverse Laplace transforms. “Given the two Laplace transforms F(s) and G(s), then L^{-1}\left\{aF(s)+bG(s)\right}=aL^{-1}\left\{F(s)\right\}+bL^{-1}\left\{G(s)\right\}.”

When trying to computer the inverse Laplace transforms, it is important to first look at the denominator and then try to identify the transform based on that. If you can’t figure it out just based on looking at the denominator, look at the numerator. Sometime you may have to manipulate the numerator to get into the correct form needed.

By looking at the table, we can see that the denominator is almost the same as the denominator of the transform for \sqrt{t}. With some algebraic manipulation to the numerator, we are able to figure out the inverse Laplace transform.

See, Laplace transforms aren’t that hard after all. They can get a little messy and can be take a long time to solve after first, but with more practice, the better you’ll get. Make sure you have that table handy!

Until next time,
Leah

## Tuesday, November 3, 2015

### Advanced Math Solutions – Ordinary Differential Equations Calculator, Exact Differential Equations

In the previous posts, we have covered three types of ordinary differential equations, (ODE). We have now reached the last type of ODE. In this post, we will talk about exact differential equations.

What is an exact differential equation?

There must be a 0 on the right side of the equation and M(x,y)dx and N(x,y)dx must be separated by a +.

Steps to solve exact differential equations:
1. Verify that \frac{∂M(x,y)}{∂y}=\frac{∂N(x,y)}{∂x}
• Find M(x,y) and N(x,y)
2. Integrate \int M(x,y)dx or \int N(x,y)dy
• This will help us find Ψ(x,y)
3. Replace c with ƞ(x) if you integrated N(x,y) with respect to y, or ƞ(y) if you integrated M(x,y) with respect to x
4. Compute ƞ(x) or ƞ(y)
5. Solve to get the implicit or explicit solution, depending on which is preferred
• Don’t forget to substitute ƞ(x) or ƞ(y)

Exact differential equations can be tricky. We will solve the first example step by step to help you better understand how to solve exact differential equations.

1. Verify that \frac{∂M(x,y)}{∂y}=\frac{∂N(x,y)}{∂x}
M(x,y)=2xy-9x^2
N(x,y)=2y+x^2+1

2. Integrate \int N(x,y)dy

3. Replace c with ƞ(x)

4. Compute ƞ(x)

We took the derivative with respect to x of Ψ(x,y), which is equal to y+x^2 y+y^2+ƞ(x). We then compared the derivative to M(x,y), the equation we didn’t integrate. We then integrated both sides to solve for ƞ(x).

5. Find the implicit equation

Exact differential equations may look scary because of the odd looking symbols and multiple steps. If you double check your work, memorize the steps, and practice, you can definitely get this concept down. Don’t be afraid and dive in!

Until next time,

Leah

## Wednesday, October 21, 2015

### Advanced Math Solutions – Ordinary Differential Equations Calculator, Bernoulli ODE

Last post, we learned about separable differential equations. In this post, we will learn about Bernoulli differential equation, which will require us to refresh our brains on linear first order differential equations. A Bernoulli differential equation is a differential equation that is written in the form:
y^'+p(x)y=q(x)y^n

where p(x) and q(x) are continuous functions on a given interval and n is a rational number. The concept of Bernoulli differential equations is to make a nonlinear differential equation into a linear differential equation. If n=0 or n=1, then the equation is linear. Bernoulli’s equation is used, when n is not equal to 0 or 1.

How to solve Bernoulli differential equations:
1. Put the differential equation in the form of Bernoulli’s equation
2. Divide by y^n
3. Put the equation in the form \frac{1}{1-n}v^'+p(x)v=q(x)
• We do this because we set v=y^{1-n} and v^'=(1-n)y^{-n}y^'
4. Solve the linear first order differential equation
• We learned how to do this in a previous post

1. Put the differential equation in the form of Bernoulli’s equation

2. Divide by y^n

3. Put the equation in the form \frac{1}{1-n}v^'+p(x)v=q(x)

4. Solve linear first order differential equation

As long as you memorize Bernoulli’s equation, the equation in Step 3, and how to solve linear first order differential equations, Bernoulli differential equations should be a piece of cake, even if they do take a long time to solve.

Fun fact: Jacob Bernoulli, who found Bernoulli’s equation, is brothers with Johann Bernoulli, who supposedly discovered L’Hopital’s rule. The Bernoulli family had many brilliant mathematicians.

Until next time,

Leah

## Tuesday, October 13, 2015

### Advanced Math Solutions – Ordinary Differential Equations Calculator, Separable ODE

Last post, we talked about linear first order differential equations. In this post, we will talk about separable differential equations. A separable differential equation is a nonlinear first order differential equation that can be written in the form:

N(y)\frac{dy}{dx}=M(x)

A separable differential equation is separable if the variables can be separated. Separable differential equations are pretty simple and do not require many steps to solve.

How to solve separable differential equations:
1. Rewrite the differential equation as N(y)dy=M(x)dx
2. Integrate both sides
• Don’t forget to add the unknown constant
3. Solve for y(x)

\frac{dy}{dx}=\frac{3x+1}{4y}

1. Rewrite the differential equation

2. Integrate both sides

3. Solve for y(x)

I think separable differential equations are the easiest ordinary differential equations. There are very few steps to solve them and they are easy to remember. Memorize the steps and you’ll be good to go!

Until next time,

Leah

## Wednesday, October 7, 2015

### Advanced Math Solutions – Ordinary Differential Equations Calculator, Linear ODE

Ordinary differential equations can be a little tricky. In a previous post, we talked about a brief overview of ODEs. In this post, we will focus on a specific type of ODE, linear first order differential equations. A linear first order differential equation is an ODE that can be put in the form
\frac{dy}{dx}+p(x)y=q(x)

where p and q are continuous functions on a given interval.  In linear first order differential equations, we can derive a formula for the solution above.

How to solve a linear first order differential equation:
1. Put the equation into form \frac{dy}{dx}+p(x)y=q(x)
• This step is very important. You cannot solve, unless the equation is in this form.
2. Set p(x)=(\ln(μ(x)))' and solve for μ(x)
• μ(x) is known as the integrating factor
• Integrate both sides
• Don’t forget the unknown constant, c, and when you can combine multiple unknown constants
3. Multiply the equation in Step 1 by μ(x) and simplify
4. Use the product rule, f\cdot g^'+f^'\cdot g on the left side of the equation to make the left side of the equation equal to (μ(x)y(x))' and write it as that:
5. Integrate and solve for y(x)
• Don’t forget the unknown constant and when you can combine multiple unknown

We will go through this first example step by step (click here for a more detailed step by step):

1. Put the equation in the form of a first order linear ODE

2. Solve for µ(x)

We integrate both sides to find µ(x).

3. Multiply by µ(x) and simplify

4. Make the left side equal to the product rule

5. Integrate and solve for y(x)

Linear first order differential equations may look intimidating at first sight. These problems do take some time to solve, but are not very hard. The most important thing to remember is to not forget about the unknown constant, c.

Until next time,

Leah

## Wednesday, September 16, 2015

### High School Math Solutions – Sequence Calculator, Sequence Examples

In the last post, we talked about sequences. In this post, we will focus on examples of different sequence problems. We will not go through them step by step, but the examples will give you a better feel for how to solve these problems.

Here is a simple example of when we use substitution to find a term (click here):

In this last example, we find the sum of the 4th term to the 100th term (click here):

Try out some of the examples of sequences we have on our site. The more you practice the better you will get!

Until next time,

Leah

## Tuesday, September 8, 2015

### High School Math Solutions – Algebra Calculator, Sequences

When dealing with simpler sequences, we can look at the sequence and get a feel for what the next term or the rule is.  However, not all sequences are nice like those. Finding the rule of a sequence can be difficult if you don’t know where to begin. We will focus on arithmetic and geometric sequences, which will make finding the rule, term, and sum of terms very easy.

Arithmetic Sequence:
• A sequence where the difference between the consecutive terms is constant
• Finding the nth term: a_n=a_1+(n-1)d
• Find the sum of n terms: S_n=\frac{n(a_1+a_n)}{2}

Geometric Sequence:
• A sequence where each term is multiplied the previous term by a constant
• Finding the nth term: a_n=a_0\cdot r^{n-1}
• Find the sum of n terms: S_n=\frac{a_1(1-r^n)}{1-r}

Here we will go through a problem involving an arithmetic sequence step by step (click here):

1. See if you can see a pattern
Find the next 3 terms of the following sequence: 3,11,19,27,35,...
We can see that difference in each term is 8.
2. Plug terms into the formula
a_n=a_1+(n-1)d
a_n=3+(n-1)8
3. Compute terms
a_6=3+(6-1)8=43
a_7=3+(7-1)8=51
a_8=3+(8-1)8=59

We will now go through a problem involving a geometric sequence step by step (click here):
Find the 7th term of the following sequence:  \frac{2}{9},\frac{2}{7},2,6,18
1. Check to see if the ratio is constant

2. Plug terms into formula
a_n=a_0∙r^(n-1)
a_n=\frac{2}{9}\cdot 3^(n-1)
3. Compute term
a_7=\frac{2}{9}\cdot 3^6=162

Here’s an example where you have to find the sum of terms (click here):

Arithmetic and geometric series aren’t very hard. Just memorize the formulas and it should be pretty easy.

Until next time,
Leah

## Saturday, August 15, 2015

### High School Math Solutions – Trigonometry Calculator, Trig Simplification

Trig simplification can be a little tricky. You are given a statement and must simplify it to its simplest form. The problem is that sometimes you don’t even know where to begin. Your eyes are glued to the problem, your head begins to hurt, and you keep writing the same equation over and over. I’ve been in that position before and hope to help make sure that never happens again. In this blog post, we will go over very helpful trig identities and some examples of trig simplification.

Here are just a couple trig identities to memorize:

Although I haven’t listed them, memorizing algebraic properties, like how to expand or factor polynomials, will be helpful too.

There is no exact formula you can follow to simplify these functions. That is why I won’t be going over a problem step by step. However, I will give you some tips on how to approach these problems.

Tips & Tricks:
1. Look for obvious simplifications when you begin
2. See if you can factor anything
3. Get rid of fractions
4. If trig functions are multiplied together, see if you can rewrite one or both
5. See what trig functions you can rewrite
6. Memorize the identities
7. Practice makes perfect
8. Keep your work nice and neat

Now that you have some of my tips and tricks in mind, let’s see some examples.

I can’t see any immediate simplifications when I begin, but I know that I can rewrite \cos^2(\theta). By doing that, it opens the function up to be simplified.

In this problem, there are two fractions. I want to get rid of them, so I combine them into one fraction. By doing that, I immediately see I can rewrite the numerator. There are now two different trig functions multiplied together in the denominator. I want to rewrite one or both of the functions, but I remember an identity and rewrite both.

Sometimes your first try of simplifying might go awry. Don’t give up! Try again. It may be frustrating, but the more practice and experience you get with these problems, the better you will become at trig simplification. Remember my tips and tricks and it should make your life a little easier.

Until next time,
Leah

## Wednesday, August 5, 2015

In our previous posts we have gone over multiple ways of solving limits. In this post we will talk about advanced limits that not only require certain methods we have talked about like, L’Hopital, squeeze theorem, etc., but also require application of properties and the algebraic limit theorem.
Here are some properties you will find very helpful when finding limits:

We won’t go each problem step by step because each problem is different from the other, but we will have two examples to show different types of advanced limits that use multiple ways to solve them.

In this example we use the algebraic property and the algebraic limit theorem to get the function in a form, where we can apply L’Hopital’s rule and find the limit.

In this example, we use the algebraic limit theorem to get the function in a form where we can apply the squeeze theorem and find the limit.

In order to solve these advanced limits it is important that the different methods and properties are fresh on your mind. Since there is no exact recipe on how to solve each advanced limit problem because each problem is different, I will tell you what I found easiest for me.