Mathematical fool needs help understanding

guymartin

Hi guys just signed up on Boards, as the title of the thread explains, I am the 'Mathematical fool' and from time to time (all the time really ) I find myself with alot of Maths related questions that I just can not answer, obviously I search the web, which does resolve alot of issues, but time and time again I have very specific questions which I can't seem to find the answers too.

Here's the thing, when I sat my Leaving Cert many moons ago, I achieved a C in, wait for it, Foundation Maths :eek: Did not like the subject at all, had no interest, never put any effort in and my result reflected this. Needless to say, the job I have worked in since has not required anything more difficult than addition and subtraction. About a year ago, for some strange reason I walked into a bookshop and bought a first year secondary school Maths book, I took it home and started to attempt a few exercises, gradually I progressed onto the ordinary Leaving Cert book and then the higher level book, my interest and love for the subject growing more and more ( exponentially :cool: ) as time went on. I am at the stage now where I think I will attempt the Leaving Cert Higher Level exam next year and possibly the Applied Maths exam also.

My problem is although I have got alot better at Maths, I still have so many questions, some very basic that I would love to get some insight into, more to increase my understanding of what it is I am actually doing than anything else.

I will post my childish questions here from time to time and possibly someone might be able to help me if they can spare the time.

Here is one to get the ball rolling:

What exactly is a function?

My understanding so far: The function y = (x)^2, means that for every value on the x-axis the corresponding value on the y-axis is the x value squared, fair enough but what exactly is a function???

Stupid question I know, but I just can't get my head around what it is in words!

bnt

Not a stupid question at all. As you say, a Function describes the relationship between variables. You could look at its as like a bit of computer programming. In your example, x is the "input" and y is the "output". For every value of x there is a corresponding value of y.

Given that y = x², then you can probably see that there is some duplication in the results: take x = -2 or x = 2, and y = 4 in both cases. So, if you take that function's inverse x = √y , and feed in y = 4, you get x = -2 and x = 2. You show this by using a ± symbol, so x = ± √y. Two inputs can give you one output, or one input can give you two outputs, depending on the function in question.

If you do a Maths-related subject in college, you'll see the y going away, to be replaced by f(x), meaning "function of x" e.g. f(x) = x² and so on. Calculus is all about functions, what they mean and what you can do with them: numbers are optional!

Yakuza

A function in the mathematical sense means something that takes one or more inputs and produces an answer (or as noted above, more than one answer).

A function basically describes the relationshop between the output ("dependent variable") and the input(s) (independent variable(s)).
Differential calculus concerns itself with how a function's output changes with respect to small changes in one of the inputs - this has a vast range of applications from statistics to finance to astrodynamics to engineering (and countless more).

Again, as noted above, you don't have to use numbers to get the meaning across - maturity is a function of age, the value of something is a function of how desirable it is and/or how rare it is, etc.

Take a look at the wiki article on functions (it's a bit stiff and formal, but worth a look IMHO)

CJC86

Just to be pedantic, a function is a relation between two sets (ie. input set and output set) such that each input has exactly one output.

So, the relation [latex]x = \pm\sqrt{y}[/latex] is equivalent to the function [latex]y=x^2[/latex], but only the second one is a function.

moycullen14

CJC86 wrote: »

Just to be pedantic, a function is a relation between two sets (ie. input set and output set) such that each input has exactly one output.

So, the relation [latex]x = \pm\sqrt{y}[/latex] is equivalent to the function [latex]y=x^2[/latex], but only the second one is a function.

Far from pedantic, IMHO.

It's a good example of the difference between a relationship and a function and, more importantly, what is and is not a function.

IIRC there was a LC question a few years ago about arctan(x) not being a function.

TheBody

The vertical line test is a useful test for a function when you have a graph. See here:

http://en.wikipedia.org/wiki/Vertical_line_test

bnt

OK - my definition of "function" wasn't sufficiently rigorous! Sorry ...

jeepers101

Some real life examples of functions:

Circles: The circumference of a circle is a function of the length of the radius.

Shadows: The length of a shadow is a function of the size of the object and the time of day.

Temperature If you've ever converted Fahrenheit to Celsius you'll have seen a relationship which can be described as a function. C= 5/9(F-32)

mawk

By the way op, good on you for taking an interest and pushing yourself to learn.
Maths is so useful and people are so scared of it :-(

Nappy

Hey OP,

Check out: http://www.khanacademy.org/

You can teach yourself nearly any aspect of mathematics from the ground up. It should have the answer to all your questions. I still use it and am in my final year of a mathematics degree.

dh0011

OP i have a degree in applied maths and almost finished my PhD. If you want to PM me at any stage about any trouble you are having you can.

guymartin

Nappy wrote: »

Hey OP,

Check out: http://www.khanacademy.org/

You can teach yourself nearly any aspect of mathematics from the ground up. It should have the answer to all your questions. I still use it and am in my final year of a mathematics degree.

Thanks, I have used this and find it very good, I also use the free Maths videos on the Engineers Ireland website

guymartin

Thanks for all the help and support so far guys.

At the moment I am struggling my way through Leaving Certificate Higher Level Trigonometry.

Here is my current problem:

sin = Opposite/Hypotenuse

Example 1:

sin30 = 0.5, 0.5 is the ratio of the opposite side to the hypotenuse. In English, the opposite side is half the magnitude of the hypotenuse at 30 degrees.

Example 2:

sin90 = 1, In English, the magnitude of the opposite side and the hypotenuse are equal at 90 degrees.

Here is the problem:

If i take 4sin30 the answer is 2. Fair enough (4)(1/2) = 2, I accept this fully. But what I can't seem to get my head around is this:

In the two examples above I could see how the answers come about graphically, the range/amplitude of the graph lies between 1 and -1. Now for the 4sin30 problem, is the range of this graph between 4 and -4, and if so would the radius of the corresponding circle be 4?

If you need me to elaborate on any particular aspect of the question, please ask, I know it is worded horribly, tried my best :-)

CJC86

guymartin wrote: »

Here is the problem:

If i take 4sin30 the answer is 2. Fair enough (4)(1/2) = 2, I accept this fully. But what I can't seem to get my head around is this:

In the two examples above I could see how the answers come about graphically, the range/amplitude of the graph lies between 1 and -1. Now for the 4sin30 problem, is the range of this graph between 4 and -4, and if so would the radius of the corresponding circle be 4?

If you need me to elaborate on any particular aspect of the question, please ask, I know it is worded horribly, tried my best :-)

By multiplying sin by 4, you are stretching the y-axis of the graph. So the range of the graph of 4sinx would be -4 and 4 as you said.

If you define a circle in polar co-ordinates, then for (x,y) on the circle [latex]x=r \cos \theta [/latex] and [latex] y=r \sin \theta [/latex], where r is the radius of the circle and [latex]\theta[/latex] is the angle. So, yes, the radius of the corresponding circle would be 4.

guymartin

CJC86 wrote: »

By multiplying sin by 4, you are stretching the y-axis of the graph. So the range of the graph of 4sinx would be -4 and 4 as you said.

If you define a circle in polar co-ordinates, then for (x,y) on the circle [latex]x=r \cos \theta [/latex] and [latex] y=r \sin \theta [/latex], where r is the radius of the circle and [latex]\theta[/latex] is the angle. So, yes, the radius of the corresponding circle would be 4.

That is perfect, very good explanation, thanks.

Dare I say it, but if in the unit circle at 30 degrees the ratio of the opposite/ hypotenuse is 1/2 ( As I said earlier, the opposite sides magnitude will be half that of the radius ) does this mean in the case of 4sin30 the hypotenuse will be 4 units in magnitude so the radius of the circle will be 4, so in this instance the magnitude of the radius will be twice that of the magnitude of the opposite side at 30 degrees, meaning the magnitude of the opposite side in this case will be 2?

If 4sin90 = 4, then what exactly does this mean, the ratio of the Opposite/Hypotenuse is 4/1, the range of the graph will be between 4 and -4, but I'm trying to picture the circle with radius 4 and at 90 degrees the opposite side is 4 times larger than the hypotenuse.
Which is not possible as at 90 degrees the opposite and hypotenuse have equal magnitude.

My problem here could be my understanding of ratios ( which has always been weak if I am being honest ), does 4/1 mean that the nominator is four times larger than the denominator? Before you shoot me please look at the title of the thread. Now shoot me :-)

locum-motion

At 90 degrees, your triangle ceases to be a triangle.
Your hypotenuse is the radius of the circle.
Your "opposite side" is co-linear (if that's a word) with the hypotenuse/radius, which is vertical.
Your "adjacent side" is 0 units long. It has disappeared.

Hope this helps.

tbh

Yakuza wrote: »

Take a look at the wiki article on functions (it's a bit stiff and formal, but worth a look IMHO)

just FYI, if you replace the "en" in that url with "simple" you get a much simpler explanation http://simple.wikipedia.org/wiki/Function_(mathematics)

Yakuza

Nice one - I never knew of the Simple English wiki, cheers!

guymartin

This one is very embarrassing but I went back to the start of the Maths book, the Algebra section.

Around 2 pages in I was looking at 'Expressions' and then I said to myself what exactly is an expression. Like I get equations and identitys but what function do expressions serve.

See attachment where I tried to be a little more specific with my question.

Yakuza

In the context of maths / algebra, an expression is a way to describe a formula or a function using numerals and symbols. It's mathematical shorthand, effectively.

You can replace "define a function whose value is found by taking three times the square of a number, adding three times the number and then adding fourteen."
with f(x) = 3x² + 3x + 14 or y = 3x² + 3x + 14.

For every x, there is a y so you should be able to graph it (you'll get a quadratic curve).

http://home.avvanta.com/~math/def2.cgi?t=expression
http://www.mathsisfun.com/definitions/expression.html

guymartin

Ok, Thanks,

I have two links below. In the first link is an expression.

In the second link I have an equation as once I plugged in values for x this equated to something.

And I could graph it as I now had x values and corresponding y values.

Correct?

Yakuza

Bingo!

Tip - there's a great site called WolframAlpha, it does all sorts of mathsy things, including graphing functions etc :

Check it out here:

guymartin

Checked out the site, looks good. Thanks for the help Yazuka

Justin1982

I think one of the simplest ways to think of a function is take an example from the real world.

Say your in a car and your pressing the accelerator (call it X) and the result of how far down you press the accelerator is the speed the car goes at (call it Y).

So the speed of the car Y is a function of the positon of the pedal X.

Function essentially describes the relationship between two related entities.

guymartin

So here is todays problem, I've gone right back to basics, any 1st year secondary school student will probably be able to help me with this one, so please feel free to do so.

Problem (See attachments):

I solved the equation:

2x+1/ 5 = 1

Result x = 2.

I then sub 2 back into the original equation and it proves x was in fact 2 because the left hand side equals the right hand side. Simple enough stuff.

Now I went onto wolfram alpha and typed the equation in just to see what the graph looks like.

Here is graphical represntation of the equation:
http://www.wolframalpha.com/input/?i=%282x%2B1%29%2F5+%3D+1

First off I noticed when x = 2, y = 1, this made sense to me as when I subbed 2 into the equation the answer was 1. So then I thought I would sub in some more values for x and see could I get out y values and things were not so straight forward, my answers did not correspond with the co-ordinates of the wolfram graph,

Reason being when I sub, lets say, 1 in for x:

2(1)+1/5 = 1

2+1/5 = 1

3/5 = 1.....................1 = 5/5

5/5 - 3/5 = 2/5

= 0.4

From the graph the co-ordinate clearly should be ( 1, 0.6 )

I can see how this could be got by, doing the following:

2x+1/ 5 = 0

2(1)+1 / 5= 0

3/5 = 0.6

Also I can see how to plot an equation like 2x+3y=4

3y = -2x+4
y = (-2/3)x + 4

To find the co-ordinates just plug in the x-values and get out the corresponding y-values, the slope will be -2/3 and it will cross the y-axis at +4, this is obvious to me because the y term is in the equation to begin with.

But when facing an equation like the original (2x+1)/5 = 1 {where there is no 'y' term present}, I dont know how to solve for y? Because it seems to me that I had to disregard the 'red 1'
2x+1/ 5 = 1

and not equate to it in order to get out the correct y-values, Why is this? Or am I just losing my marbles altogether.

guymartin

Yakuza wrote: »

Bingo!

Tip - there's a great site called WolframAlpha, it does all sorts of mathsy things, including graphing functions etc :

Check it out here:

Thanks for the WolframAlpha link, I've been on it all day. Anyone got any experience using the full version?

guymartin

Justin1982 wrote: »

I think one of the simplest ways to think of a function is take an example from the real world.

Say your in a car and your pressing the accelerator (call it X) and the result of how far down you press the accelerator is the speed the car goes at (call it Y).

So the speed of the car Y is a function of the positon of the pedal X.

Function essentially describes the relationship between two related entities.

So a function in lay mans terms is a operation you perform on a variable to get a result.

x is a variable,
squaring, cubeing etc... is the function
and y is the result of the function.

Is this why we say f(x) = y, because the result y is some function ( an action we perform on the variable x) of x??

God I hope I have it this time or i'll never get it:D

locum-motion

There's only one variable in

2x+1/ 5 = 1. (Which I suppose means there's no variable at all; x can't vary here, it's 2 and only 2 and that's it)

You can't graph one variable. To graph something, it has to have an x and a y axis, along the lines of 'if x is something, what is y?'

If your equation was

2x+1/ 5 = y,

Then you could graph it:
When x is 1, y is 0.6
When x is 2, y is 1
When x is 3, y is 1.4
When x is 4, y is 1.8
When x is 5, y is 2.2
Etc.

Now you have a set of points that you can plot on a graph.

When you tried to substitute other values for x in your original equation, WA said 'false' because it WAS false: (2*3)+1 /5 will never = 1, no matter how many times you try to shoehorn it in!

This is my "laymans' language" explanation. As I've said before on this forum, I don't have a high level of formal maths education, so I apologise if I've used incorrect terminology.

guymartin wrote: »

So here is todays problem, I've gone right back to basics, any 1st year secondary school student will probably be able to help me with this one, so please feel free to do so.

Problem (See attachments):

I solved the equation:

2x+1/ 5 = 1

Result x = 2.

I then sub 2 back into the original equation and it proves x was in fact 2 because the left hand side equals the right hand side. Simple enough stuff.

Now I went onto wolfram alpha and typed the equation in just to see what the graph looks like.

Here is graphical represntation of the equation:
http://www.wolframalpha.com/input/?i=%282x%2B1%29%2F5+%3D+1

First off I noticed when x = 2, y = 1, this made sense to me as when I subbed 2 into the equation the answer was 1. So then I thought I would sub in some more values for x and see could I get out y values and things were not so straight forward, my answers did not correspond with the co-ordinates of the wolfram graph,

Reason being when I sub, lets say, 1 in for x:

2(1)+1/5 = 1

2+1/5 = 1

3/5 = 1.....................1 = 5/5

5/5 - 3/5 = 2/5

= 0.4

From the graph the co-ordinate clearly should be ( 1, 0.6 )

I can see how this could be got by, doing the following:

2x+1/ 5 = 0

2(1)+1 / 5= 0

3/5 = 0.6

Also I can see how to plot an equation like 2x+3y=4

3y = -2x+4
y = (-2/3)x + 4

To find the co-ordinates just plug in the x-values and get out the corresponding y-values, the slope will be -2/3 and it will cross the y-axis at +4, this is obvious to me because the y term is in the equation to begin with.

But when facing an equation like the original (2x+1)/5 = 1 {where there is no 'y' term present}, I dont know how to solve for y? Because it seems to me that I had to disregard the 'red 1'
2x+1/ 5 = 1

and not equate to it in order to get out the correct y-values, Why is this? Or am I just losing my marbles altogether.

locum-motion

As far as I understand it, you're right in this post (no 27).
Now, take that knowledge back to what you posted in no 25, and look at it again. Perhaps use my post (28) as a lens to look at 25 again, and you should grasp it.

guymartin wrote: »

So a function in lay mans terms is a operation you perform on a variable to get a result.

x is a variable,
squaring, cubeing etc... is the function
and y is the result of the function.

Is this why we say f(x) = y, because the result y is some function ( an action we perform on the variable x) of x??

God I hope I have it this time or i'll never get it:D

locum-motion

I think that what you've graphed here in your second link is
x^2 - 6x + 5 = y,
not
x^2 - 6x + 5 = 0.

guymartin wrote: »

Ok, Thanks,

I have two links below. In the first link is an expression.

In the second link I have an equation as once I plugged in values for x this equated to something.

And I could graph it as I now had x values and corresponding y values.

Correct?

guymartin

I can see why you would need to have a 'y' term in an equation to, so you could let all the other terms in the equation be equal to 'y' and then get out your (x,y) values.

But if this is the case then look at the attachment '1&2' below, if I have the equation x^2+6x+5=0 and I sub in values for 'x' I get the 'y' values out. See the graph at this link: http://www.wolframalpha.com/input/?i=x%5E2%2B6x%2B5%3D0

Now I can make sense of this because we are performing an 'action/function' on x, namely we are squaring 'x', adding 6 times 'x' then adding 5 to it, as we are performing an action on 'x' the result is 'y'.

But this kind of goes against my interpretation of what you were saying to me in post #28 as there was no 'y' term in the equation to begin with yet I can still obtain a 'y' value.

My interpretation of what is happening is that the function of 'x' gives us 'y', there does not have to be a 'y' term in the equation in order to get 'y' as the function of 'x' is 'y'.

If I take a second look at the equation (2x+1) / 5 = 1, In English what I see is this:

Here is a function of 'x' that gives you 'y' now what must 'x' be in order for the output to be 1.

You were right in post #28, 'x' is not a variable as 2 and only 2 can be the answer.

I think I have just answered my own question while writing this. Take a look at this graph: http://www.wolframalpha.com/input/?i=%282x%2B1%29%2F5+%3D+1

What this graph is saying to me in English is this: when 'x' is equal to 2 and we perform the function (2x+1)/5 to 2 the answer is 1. But if we want to make a line out of this equation we need to sub in values for x into the function (2x+1)/5 and the result will be the 'y' value. 1 has nothing to do with any other co-ordinates on the line, it is just the y co-ordinate when 'x' is equal to 2.

So what I have done in the attachment named 'Equation 1 Alternative Method' is altered the function, so now when I put in values for 'x' I get out corresponding 'y' values but these are 'y' for a different line to the line (2x+1)/5 = 1

MathsManiac

This might help:

Expressions, equations, and functions are not the same thing.

x^2 - 5x + 4 is an expression.
x^2 - 5x + 4 = 0 is an equation.
y = x^2 - 5x + 4 is also an equation. One like this can also be taken to define a function (provided it's clear what the domain and codomain are - these are terms you can look up.)

x^2 + 6x + 1 is an expression in one unknown.
x^2 +2xy + y^2 is an expression in two unknowns.
xyz/(x+y+z) is an expression in three unknowns, etc.
5 is an expression in 0 unknowns - such an expression is called a constant expression, or just a constant.

An equation is a statement that two expressions are equal.

If you have an equation in one unknown, then a value of the unknown that makes the equation true is called a solution of the equation.
If you have an equation in two unknowns, then a pair of values for the unknowns that makes the equation true is a solution.
If you have an equation in three unknowns, then a triple of values for the unknowns that makes the equation true is a solution, etc.

e.g.
x=1 is a solution of the equation x^2 - 5x + 4 = 0.
x=2 is not a solution of the equation x^2 - 5x + 4 = 0.
(x, y) = (2, -2) is a solution of the equation y = x^2 - 5x + 4.

Equations can have lots of solutions, or they can have none.

An equation which is true for all possible values of the variables concerned is called an identity.

For example: x^2 + 2x = x(x + 2) is an identity.

You cannot "solve" an expression. You can only solve an equation. This means finding all the solutions of the equation. The instruction "Solve x^2 - 5x + 4 = 0" is meaningful. The instruction: "Solve x^2 - 5x + 4" is meaningless. However, equations where one side is 0 are so common, that if you ask a computer algebra package to "Solve x^2 - 5x + 4", it will probably assume you meant "Solve x^2 - 5x + 4 = 0" and do that.

If you have an equation in two unknowns, then this defines a relation between those unknowns. The relation is the set of all couples that are solutions of the equation. If you draw a diagram showing all of these couples, this is called the graph of the relation.

If you have a relation between x and y, then you might or might not have a function. That is, y might or might not be a function of x. In order to be a function, each x-value can only be associated with one y-value. (If you look at the graph of a function, any vertical line cannot cross it more than once.)

In the relation x^2 + y^2 = 4, where x and y are real numbers, y is not a function of x, (because, for example, the couple (0,2) and (0,-2) are both solutions - two different values of y with the same value of x).
In the relation y = x^2 - 5x + 4, where x and y are real numbers, y is a function of x, as each value of x produces only one value of y.

There are quite a lot of different mathematical terms there, but I hope that helps to distinguish clearly between them. A lot of people don't retain this level of precision in their ordinary work.

locum-motion

MathsManiac wrote: »

...

An equation which is true for all possible values of the variables concerned is called an identity.

For example: x^2 + 2x = x(x + 2) is an identity.
...

Thanks. I was wondering what the word meant when you used it on the other thread the other day.

locum-motion

MathsManiac wrote: »

This might help:

Expressions, equations, and functions are not the same thing.

x^2 - 5x + 4 is an expression.
x^2 - 5x + 4 = 0 is an equation.
y = x^2 - 5x + 4 is also an equation...

That's what I was trying to get at the other evening. I finished my post with the following:

locum-motion wrote: »

...
This is my "laymans' language" explanation. As I've said before on this forum, I don't have a high level of formal maths education, so I apologise if I've used incorrect terminology.

Thanks for the clarifications.