How Does Calculus Help Solve Patterns?

[lenfromkits]

I need to show that this pattern is true.

\sqrt{n+1} \approx \frac{n}{2} + 1 \ \ when \ n << 1

the smaller 'n' is, the more this is true.

(i.e. for example, \sqrt{1.0000000000001234} \approx \frac{.0000000000001234}{2} + 1 )

Can anyone help?

(this is not a homework assignment. I need it for work)

Thanks.

 

[Matterwave]

You can do a Taylor expand around 0 of \sqrt{x+1}

It is:
\sqrt{x+1}=1+x\left[\frac{1}{2\sqrt{x+1}}\right]_{x=0}+...

 

[elfmotat]

The linear approximation of f(n) about n_0 is given by:

L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)If f(n)=\sqrt{n+1}, then the linear approximation of f(n) around 0 is:

L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1

 

[lenfromkits]

Thanks! I can see that you have solved my problem. Now I just need to understand 90% more than I understand now :)

Can I please ask where this comes from:

L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)

Thanks,
Len

The linear approximation of f(n) about n_0 is given by:

L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)


If f(n)=\sqrt{n+1}, then the linear approximation of f(n) around 0 is:

L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1

 

[Curl]

It's called linear approximation. Approximate your function with a line whose slope is the same as the original function's slope at x=0.
You find this slope from the derivative.

 

[willem2]

Even simpler, since

n + 1 \approx n + 1 + \frac {n^2} {4} = (n+\frac{1}{2})^2

if n<<1


\sqrt {n+1} \approx n + \frac{1}{2}

 

[NascentOxygen]

You could apply the Binomial Theorem. After the second term, subsequent terms become vanishingly small. http://en.wikipedia.org/wiki/Binomial_series

 

[elfmotat]

Thanks! I can see that you have solved my problem. Now I just need to understand 90% more than I understand now :)

Can I please ask where this comes from:

L(n)=f&#039;(n_0)(n-n_0)+f(n_0) \approx f(n)

Thanks,
Len

You use a tangent line to approximate the function at n_0. The equation of the tangent line is obviously given by y=f&#039;(x_0)(x-x_0)+f(x_0).

http://www.function-grapher.com/graphs/fc544b3a572de466f056ae4620193c3f.png