In this post I’ll be proving to you that:

Now firstly I will have to say that:

And also that:

If this is the case, then…

Since this is in the form:

I would have to conclude that:

Hence I’ve proven that:

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:

And expanded, it looks like this:

[*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 |
---|---|

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

Now, let’s replace

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

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:

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 |
---|

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

Since:

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

This means that:

And finally, when x=π:

This is because:

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

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