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**e^x**

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

How would we expand e^x as an infinite series ?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

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.

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

How does exponential theorem for exponential series works?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

hi 666 bro

Not sure what that is. Do you mean Maclaurin's theorem?

You assume

set x = 0 and the equation becomes 1 = A.

Now differentiate

set x = 0 and we have 1 = B

Differentiate again

set x = 0 and we have C = 1/2

Carrying on we get D = 1/6 … E = 1/24 and the general term is 1/r!

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

Is gradient of line means slope ?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 44,425

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

What does Δx means in graph of f(x) = a^x ?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

hi 666 bro,

Are you asking about post 2 ?

I was finding out dy/dx for the graph using basic calculus.

When differential calculus was invented the process for finding out a gradient function was developed. It goes like this:

Pick a general point (x,y) on the graph. And a second point close to the first.

Calculate algebraically the gradient of the line joining those points.

Analyse what happens to this gradient as the second point moves closer to the first. If a clear limit emerges that is the gradient function.

One way of showing that the second point is close to the first is to use 'delta' notation. Δx means a little bit in the x direction. This is explained very well here:

https://www.mathsisfun.com/calculus/introduction.html

Δ is a capital letter from the Greek alphabet. The small version of the same letter is:

When I pick the symbol from the ones at the top of the MIF page, only a capital is available. When I use Latex to get mathematical notation the command \delta gives small letter. So I had to switch between the two in post 2. Sorry about that. Perhaps someone knows how to get a small delta when using text only or a capital when using Latex. Please let us know if you do.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

You've said that "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"

So, my question is how could we see a gets bigger and bigger visually and please explain with an example?

*Last edited by 666 bro (2020-05-11 02:13:30)*

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

It seems that you haven't done one of the things I suggested.

Go to this page:

https://www.mathsisfun.com/data/function-grapher.php

On the top (blue) line enter (y=) 2^x

On the lower (orange) line enter 3^x

You will get two curves with similar shape. Both go through (0,1)

Try varying the 3, say try 4^x.

You'll see the effect that has on the curve.

Experiment with different values so you are trying y = a^x for many different values of a.

You should see that the bigger 'a' is, the faster the curve rises when x>0 and also the faster it drops when x<0.

Make the 'orange' graph into (y=) x + 1 so you have a straight line, gradient 1, going through (0,1)

Adjust the blue 'a' value to 2.1.... until the straight line looks like it is a tangent to the curve.

If you try y = e^x you'll see it's a perfect tangent.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

In that post 8 you've said that you're finding out dy/dx for the graph by taking general point (x,y) and second point close to first one and calculate slope of line joining those points. I've tried for the functions like 2^x , 3^x but what I'm confused about is had you calculated slope of the line or derivative as a limit?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

Post 2 has the first part of a series of lessons that will lead to calculating dy/dx for any a^x. The steps to achieve this are (in outline only)

(1) Get a limit expression for the gradient at (0,1) but we cannot evaluate the limit at this stage. Call it k anyway even though we don't know what it is yet.

(2) Get a limit expression for the gradient at any point. Once again cannot evaluate it at this stage.

(3) Note that for y = a^x the gradient function dy/dx is ky.

So if we can find k, we can differentiate any a^x.

(4) Investigate the particular value of a for which k=1. Call this e.

(5) So when y = e^x, dy/dx = e^x

(6) Find a series expansion for e^x. [By putting x = 1 we can evaluate e.]

(7) Show that all functions that differentiate to give themselves differ only by a multiplier.

That's as far as post 2 goes. Still to come if you want it:

(8) Find how to differentiate y = ln(x) where ln is the natural logarithm, ie base e.

(9) Hence integrate 1/x.

(10) Use change of log base rules to differentiate y = log(x) when the log is in any base.

(11) Hence determine k for any a.

Post back if you would like to see these remaining steps.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 706

So, which concepts are required to understand this series?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

algebra

limits

differentiation

integration

graphs of 'families' of similar functions

rules of logarithms

B

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathland****Member**- Registered: 2021-03-25
- Posts: 444

Bob wrote:

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

Continue with your notes. Interesting so far.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,863

As a teacher with many years of experience I know that the starting point in any lesson is to check that the pupils have understood the previous lesson. From my point of view, there is no point spending my time adding the next steps if you haven't got a good grasp of what I have said so far. I'm surprised you feel ready for the next step so soon after getting the link to the thread.

So I want something from you first:

Using a graph plotter, try a = 4 then 5 then 1 then 1/2 then 1/3.

Comment on what you are observing in these cases. Make three conclusions about how the value of a affects the graph.

Then use the series expansion for e^x with x = 1 to compute e to 8 decimal places .

Post the steps in your working and your final result. Pay particular attention to explaining how you know your answer is accurate to 8 dp (without just comparing with a published result).

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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