Math Is Fun Forum

The measurements given in the question are distances, not 'points' (exact locations) as such, and the error is in their placement within the 'points' notation (ie, the x,y elements in the brackets).

yes.

hi 666 bro

I know you're trying to self-study so I'll give you the lesson I have used in the past for my A level students. It'll be longer than just answering your question, but I think you'll find it covers lots of topics that you are interested in. For my class it would take more than one lesson to do all this and, of course, if I said something a student didn't follow, they could ask questions and get me to go back over earlier topics.  So take your time; do as much as you can in one session; and post back if you have any questions.

The lesson starts with "What does the graph of y = 2^x look like? I suggest you try this.  Make up a table of values for y when x = -3, -2, -1, 0, 1, 2, 3 and plot the points.

Then repeat for y = 3^x and consider other values of graphs of the form y = a^x.  I'll let you try this and leave a big space before adding my graphs.

https://i.imgur.com/W3ycKiy.gif

Graphs of the form y = a^x form a 'family' of similar curves.  They all have a similar shape and all go through (0,1).  As 'a' gets bigger the graphs go up more steeply

Next step.  What is the gradient at (0,1)

Form an expression for the chord joining (0,1) to (Δx , a^Δx) and let  Δx tend to 0.

Now, at this stage, we've got no way to say what 'k' is but later we'll find out.  One thing should be clear; each value of 'a' will have a limit because we can see that the gradient at (0,1) exists, and is different for each 'a' value, and gets bigger as 'a' gets bigger.

Now the general gradient at any point on the curve.

So this family of curves has a very interesting property.  The gradient function for each is the function again multiplied by the gradient at (0,1).

Next step.

So can we find an 'a' so that k = 1.  This would mean that function differentiates to give itself!

Looking at the graphs we can see that different 'a' values give different 'k' values, so let's assume that such a value of 'a' exists.  As it's special I'll call that one 'e' rather than 'a'.  In other words

So e^x is a function that differentiates to give itself.   But is it the only function with this property?

Let's say there are two such functions, f(x) and g(x).  So df/dx = f and dg/dx = g.

Consider the function h = f/g and differentiate it.  This needs the quotient rule.

If h differentiates to zero then h is a constant so f is a multiple of g.  There are many functions that differentiate to give themselves back, but they're all multiples of each other.

Next step.  So can we find a function that just has powers of x that differentiates to give itself back.  (This is the answer to your question!)

I tell the class that the function has a '1' in it so it looks like this:

But what differentiates to give that 1?  Answer x

So the function must look like this:

But what differentiates to give that x ? Answer x^2/2

So the function must look like this:

But what diffentiates to give x^2/2 ? Answer x^3 / 6

So the function must look like this:

Hopefully one of my class now says "That means it goes on for ever!".  It's always great when they make their own discoveries like this.

So we carry on adding terms and also find an expression for the general term:

As many function exist that differentiate to give themselves this doesn't yet prove that e^x and that power series are the same but that'll come later.  Looking at the a^x graphs it certainly comes at the right place in the family because you can work out the gradient for 2^x and it's under 1; and the gradient for 3^x and it's over 1, so our special 'e' number must be between 2 and 3.  If you put x = 1 in the power series you'll get the value of 'e' and it lies in that range.

There is more but I'll take a break for now,

Bob

hi mathland

Somehow I knew in advance that you would want to know about this so I searched back through all my posts and found this:

http://www.mathisfunforum.com/viewtopic.php?id=25550

My last post in the thread was an offer to finish the lessons but only if the guy was interested.  I got no response then, but the offer still stands.

Reply on that thread please rather than this one.

Bob

Now, I'm highly got interested in learning geometry by playing a game called "euclidea" but I forgot all the basics that I've learnt in my school so, where should I start learning?
Any  resources and suggestions always welcome.

Know that -a^2 means -1 x a^2 and not -a x -a.
The reason is that -a^2 has its coefficient as -1 
Therefore -a x -a = a^2 and and not -a^2.
This symbol ^ implies exponent.

It might help to consider what we actually mean by e^x.

What is the definition of e^x?

Further notes on this:

cuberoot(t+4) is cuberoot(t) translated 4 units left in a horizontal direction.

Examine y = (x)^(1/3)

This graph and its inverse are 1:1 functions  so we can proceed  to raise both sides to the power 3

y^3 = x

We can reflect the graph of this function in the line y=x by swapping the x and y terms.

y = x^3

We know all about this cubic already.  Both its domain and range are (-∞,∞)

The reflection interchanges  domain and range.  In other examples this might matter but as both are  (-∞,∞) it doesn't matter.

The translation might effect some domains, but again, not this one.

So we can get domain and range just by considering simple transformations of the graph.

General hint.  Cubics in general have certain common properties. We can call all such curves a 'family of curves' and use general properties to learn about new members.

Cubics with a positive x^3 term all start in the 3rd quadrant and finish in the 1st. If x^3 is negative this becomes 2nd and 4th quadrants.

Cubics have two stationary points (max and min) unless the two points merge into one in which case we have a point of inflexion.

Here are some general properties for quadratics:

All have a line of symmetry  x = -b/2a .  This line goes through the stationary point which is a minimum when the x^2 term is positive and a maximum when the x^2 term is negative.

Graphs of the form y = a^x

These all go through (0,1)

The gradient function is k . a^x

When k = 1, we use 'e' , making a function that has the interesting property that dy/dx = y.

I'm sure I did a thread on this but I cannot find it now using search.

Bob