# 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.

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…

[*Image can be seen hereÂ 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…

What 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.

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

Related:

How to derive Euler’s Identity using the Maclaurin Series

# How to differentiate y=arctanx

Below I’m going to demonstrate how to integrate y=arctanx…

Firstly, we need to know that:

$\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1\\ \\ \frac { \sin ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } +\frac { \cos ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } =\frac { 1 }{ \cos ^{ 2 }{ y } } \\ \\ \tan ^{ 2 }{ y } +1=\sec ^{ 2 }{ y }$

We also need to know that:

$x=\tan { y } \\ \\ x=\frac { \sin { y } }{ \cos { y } } \\ \\ x\cdot \cos { y } =\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } +x\cdot \left( -\sin { y } \right) =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } -x\sin { y } =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +x\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ \frac { 1 }{ \cos { y } } \cdot \frac { dx }{ dy } \cdot \cos { y } =\frac { 1 }{ \cos { y } } \left( \cos { y+\frac { \sin ^{ 2 }{ y } }{ \cos { y } } } \right) \\ \\ \frac { dx }{ dy } =1+\tan ^{ 2 }{ y } \\ \\ \therefore \quad \frac { dx }{ dy } =\sec ^{ 2 }{ y }$

And finally:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Now, using implicit differentiation:

$y=\arctan { x } \\ \\ \tan { y } =x\\ \\ \sec ^{ 2 }{ y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \sec ^{ 2 }{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \tan ^{ 2 }{ y+1 } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ { x }^{ 2 }+1 }$

# Logarithmic Differentiation As Seen In Video…

In this video, Patrick JMT demonstrated how to do logarithmic differentiation.

Now I’m going to show you how I’d solve the same problem…

Firstly we must know that:

$y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ If\quad y=u\cdot v,\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ u=x,\quad \frac { du }{ dx } =1\\ \\ v=\ln { \left( \ln { x } \right) } =\ln { q } \\ \\ \frac { dv }{ dq } =\frac { 1 }{ q } =\frac { 1 }{ \ln { x } } \\ \\ q=\ln { x } ,\quad \frac { dq }{ dx } =\frac { 1 }{ x } \\ \\ \therefore \quad \frac { dv }{ dx } =\frac { 1 }{ x\ln { x } } \\ \\ \therefore \quad \frac { dy }{ dx } =x\cdot \frac { 1 }{ x\ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) }$

Now:

$y={ \left( \ln { x } \right) }^{ x }\\ \\ \ln { y=\ln { \left( { \left( \ln { x } \right) }^{ x } \right) } } \\ \\ \ln { y } =x\ln { \left( \ln { x } \right) } \\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\} \\ \\ \frac { dy }{ dx } ={ \left( \ln { x } \right) }^{ x }\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\}$

# Derivative – 27/02/15

$n\ln { x } =y\\ \\ \ln { \left( { x }^{ n } \right) } =y\\ \\ \log _{ e }{ \left( { x }^{ n } \right) } =y\\ \\ { e }^{ y }={ x }^{ n }\\ \\ { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\\ \\ \frac { 1 }{ { e }^{ y } } \cdot { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\cdot \frac { 1 }{ { e }^{ y } } \\ \\ \frac { dy }{ dx } =n{ x }^{ n-1 }\cdot { x }^{ -n }\\ \\ \frac { dy }{ dx } =n{ x }^{ n-1+\left( -n \right) }\\ \\ \frac { dy }{ dx } =n{ x }^{ -1 }\\ \\ \frac { dy }{ dx } =\frac { n }{ x }$

# Anti-Derivative Proof – 27/02/15

$y=\frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \ln { y } =\ln { \left( \frac { { x }^{ \left( n+1 \right) } }{ \left( n+1 \right) } \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } -\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } +{ C }_{ 2 }\\ \\ \therefore \quad { C }_{ 2 }=-\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\left( n+1 \right) \cdot \ln { x } +{ C }_{ 2 }\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \cdot y\\ \\ \frac { dy }{ dx } =\left( n+1 \right) \cdot { x }^{ -1 }\cdot \frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1 }\cdot { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1+\left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ n }\\ \\ \therefore \quad \int { { x }^{ n } } dx=\frac { 1 }{ \left( n+1 \right) } \cdot { x }^{ n+1 }+C$

# Integration Problem – Finding the area of quarter of a circle by integrating…

Integrate:

$\int _{ 0 }^{ r }{ \sqrt { { r }^{ 2 }-{ x }^{ 2 } } } dx$

Say that:

$x=r\sin { \theta }$

So…

$x=r\sin { \theta } \\ \\ x=ru\\ \\ \frac { dx }{ du } =r\\ \\ u=\sin { \theta } \\ \\ \frac { du }{ d\theta } =\cos { \theta } \\ \\ \therefore \quad \frac { dx }{ d\theta } =r\cos { \theta } \\ \\ \therefore \quad dx=r\cos { \theta } d\theta$

Now…

$When\quad x=r,\\ \\ r\sin { \theta =r } \\ \\ \sin { \theta } =1\\ \\ \therefore \quad \theta =\frac { \pi }{ 2 } \\ \\ As\quad \left\{ 0\le \theta \le \frac { \pi }{ 2 } \right\} \\ \\ When\quad x=0,\\ \\ r\sin { \theta } =0\\ \\ \sin { \theta } =0\\ \\ \therefore \quad \theta =0$

So you now have to integrate:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }-{ { r }^{ 2 }\sin ^{ 2 }{ \theta } } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\cos ^{ 2 }{ \theta } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( { r }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } { \left( \cos ^{ 2 }{ \theta } \right) }^{ \frac { 1 }{ 2 } }\cdot r\cos { \theta } d\theta \\ \\ =\int _{ o }^{ \frac { \pi }{ 2 } }{ r\cos { \theta } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \cos ^{ 2 }{ \theta } d\theta$

But first, realise that:

$\cos { \left( \theta +\theta \right) } =\cos { \theta } \cos { \theta } -\sin { \theta } \sin { \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -1+\cos ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1\\ \\ 2\cos ^{ 2 }{ \theta } =\cos { 2\theta } +1\\ \\ \cos ^{ 2 }{ \theta } =\frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 }$

So you now integrate:

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \left( \frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 } \right) d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } { r }^{ 2 }\cos { 2\theta } +\frac { 1 }{ 2 } { r }^{ 2 }\quad d\theta$

And:

$If\quad p=\frac { 1 }{ 4 } { r }^{ 2 }\sin { 2\theta } =\frac { 1 }{ 4 } { r }^{ 2 }u\\ \\ \frac { dp }{ du } =\frac { 1 }{ 4 } { r }^{ 2 }\\ \\ u=\sin { 2\theta } =\sin { q } \\ \\ \frac { du }{ dq } =\cos { q } =\cos { 2\theta } \\ \\ q=2\theta ,\quad \frac { dq }{ d\theta } =2\\ \\ \frac { du }{ d\theta } =2\cos { 2\theta } \\ \\ \therefore \quad \frac { dp }{ d\theta } =\frac { 1 }{ 4 } { r }^{ 2 }\cdot 2\cos { 2\theta } \\ \\ =\frac { 1 }{ 2 } { r }^{ 2 }\cos { 2\theta }$

Therefore:

Now, to get the area of a circle, you multiply the final result by 4.

$4\cdot \frac { 1 }{ 4 } \pi { r }^{ 2 }\\ \\ =\pi { r }^{ 2 }\\ \\ \therefore \quad A=\pi { r }^{ 2 }$

# Implicit Differentiation Rules

Rule 1:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Rule 2:

# If y=ln(f(x)), dy/dx=(f'(x))/f(x)

Prove that if:

$y=\ln { \left( f\left( x \right) \right) } ,\quad \frac { dy }{ dx } =\frac { f$

Firstly, say that:

$y=h\left( x \right) \\ \\ g=f\left( x \right)$

So: