How to derive Euler’s Identity using the Maclaurin Series

In this post I’ll be showing you how to derive Euler’s identity using the Maclaurin Series. It turns out that the Maclaurin series looks like this:

maclaurin series

And expanded, it looks like this:

maclaurin expansion

[*A larger version of this image can be found here.]

Now, since we want to derive Euler’s identity, we first have to find out what the formula for e^x looks like. In order to get this formula we must use the table below:

Derivatives of e^x When x=0
{ f }^{ \left( 0 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 0 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 1 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 1 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 2 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 2 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 3 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 3 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 4 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 4 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 5 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 5 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 6 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 6 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 7 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 7 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 8 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 8 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 9 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 9 \right) }\left( 0 \right) ={ e }^{ 0 }=1
{ f }^{ \left( 10 \right) }\left( x \right) ={ e }^{ x } { f }^{ \left( 10 \right) }\left( 0 \right) ={ e }^{ 0 }=1

Ok. So we’ve got a useful table just above. Let’s write out the function of e^x in its Maclaurin Series form:

f\left( x \right) ={ e }^{ x }\\ \\ =\frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( 0 \right) { x }^{ 0 }+\frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( 0 \right) { x }^{ 1 }+\frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( 0 \right) { x }^{ 2 }\\ \\ +\frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( 0 \right) { x }^{ 3 }+\frac { 1 }{ 4! } { f }^{ \left( 4 \right) }\left( 0 \right) { x }^{ 4 }+\frac { 1 }{ 5! } { f }^{ \left( 5 \right) }\left( 0 \right) { x }^{ 5 }\\ \\ +\frac { 1 }{ 6! } { f }^{ \left( 6 \right) }\left( 0 \right) { x }^{ 6 }+\frac { 1 }{ 7! } f^{ \left( 7 \right) }\left( 0 \right) { x }^{ 7 }+\frac { 1 }{ 8! } { f }^{ \left( 8 \right) }\left( 0 \right) { x }^{ 8 }\\ \\ +\frac { 1 }{ 9! } { f }^{ \left( 9 \right) }\left( 0 \right) { x }^{ 9 }+\frac { 1 }{ 10! } { f }^{ \left( 10 \right) }\left( 0 \right) { x }^{ 10 }+...

Now, let’s replace

values

 

 

with the values from the table. If we do this, the formula for e^x will become:

e^x formula

Alright, so far, so good. We are certainly on the right track. Our next goal will be to discover what e^(i*x) is. This is because to produce Euler’s identity, we need to come up with:

{ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }

To come up with the formula above, we will need the table below, because our latest e^(x) formula will have to be transformed. x will be turned into i*x.

Imaginary Numbers Exponentiated
\sqrt { -1 } =i
i\cdot i={ i }^{ 2 }=-1
{ i }^{ 3 }={ i }^{ 2 }\cdot i=-i
{ i }^{ 4 }={ i }^{ 3 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1
{ i }^{ 5 }={ i }^{ 4 }\cdot i=i
{ i }^{ 6 }={ i }^{ 5 }\cdot i=i\cdot i={ i }^{ 2 }=-1
{ i }^{ 7 }={ i }^{ 6 }\cdot i=-i
{ i }^{ 8 }={ i }^{ 7 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1
{ i }^{ 9 }={ i }^{ 8 }\cdot i=i
{ i }^{ 10 }={ i }^{ 9 }\cdot i={ i }^{ 2 }=-1
{ i }^{ 11 }={ i }^{ 10 }\cdot i=-i

As we’ve got the table above, we can figure out what the formula e^(i*x) would look like:

e^(ix) expanded

Since:

cosx sinx

[*To find out why it’s the case, visit this page.]

This means that:

{ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }

And finally, when x=π:

{ e }^{ i\pi }=\cos { \left( \pi \right) } +i\sin { \left( \pi \right) } \\ \\ \therefore \quad { e }^{ i\pi }=-1\\ \\ \therefore \quad { e }^{ i\pi }+1=0

This is because:

\cos { \left( \pi \right) } =-1\\ \\ \sin { \left( \pi \right) } =0

You have produced Euler’s identity from almost absolute scratch. Give yourself a pat on the back! 🙂


Related:

Deriving the Taylor Series from scratch

Deriving the Taylor Series from scratch

[Please note: In order to derive the Taylor Series, you will need to understand how to differentiate. If you know how to differentiate, finding the Taylor Series won’t be much of a problem. You also need to know that 0!=1, 1!=1, 2!=2, 3!=6, x^0=1, x^1=x.]

In this post I will be demonstrating how one can produce the Taylor Series from absolute scratch.

taylor series function

First of all, let’s look at the diagram above. Now, let’s suppose that the equation of the function above is:

f\left( x+a \right) ={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...

Ok, so we have the equation for the function, however, it isn’t complete. C_0, C_1, C_2, C_3 etc are hidden constants. This means that our second task will be to discover these constants. We need to discover these constants to find the complete equation of the function so that we can arrive at the Taylor Series. Fortunately, this task won’t be too difficult. Let me show you how C_0, C_1, C_2, C_3 etc can be found fairly easily…

When x=0:

{ f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }\quad \therefore \quad \frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }

Now:

{ f }^{ \left( 1 \right) }\left( x+a \right) ={ C }_{ 1 }+2{ C }_{ 2 }x+3{ C }_{ 3 }{ x }^{ 2 }+...

When x=0:

{ f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }\quad \therefore \quad \frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }

Also:

{ f }^{ \left( 2 \right) }\left( x+a \right) =2{ C }_{ 2 }+6{ C }_{ 3 }x+...

When x=0:

{ f }^{ \left( 2 \right) }\left( a \right) =2{ C }_{ 2 }\quad \therefore \quad \frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( a \right) ={ C }_{ 2 }

And, finally:

{ f }^{ \left( 3 \right) }\left( x+a \right) =6{ C }_{ 3 }+...

When x=0:

{ f }^{ \left( 3 \right) }\left( a \right) =6{ C }_{ 3 }\quad \therefore \quad \frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( a \right) ={ C }_{ 3 }

Alright, so now that we have discovered the hidden constants C_0, C_1, C_2 and C_3, our third task is to write down the complete equation of the function f(x+a). Thanks to the information we have above, the fact that x^0=1 and x^1=x, plus our ability to spot patterns, we will be able to do this quite quickly…

taylor series (1)

[*Image can be seen here if it appears to be too small on this page.]

And it turns out that the equation we have just above is the Taylor Series function. It can be simplified to look like this…

taylor seriesWhat is also interesting is that if we transform a=0, we get the Maclaurin Series function which can be used to discover formulas for things such as e^x.

maclaurin series

If you have any questions regarding this post, please leave your comments below. Once again, thanks for stopping by! 🙂


Related:

How to derive Euler’s Identity using the Maclaurin Series

Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

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Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

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