Can calculus solve this?

  • #1
107
0
I need to show that this pattern is true.

[itex]\sqrt{n+1} \approx \frac{n}{2} + 1 \ \ when \ n << 1[/itex]

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

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

Can anyone help?

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

Thanks.
 
Last edited:

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
You can do a Taylor expand around 0 of [itex]\sqrt{x+1}[/itex]

It is:
[tex]\sqrt{x+1}=1+x\left[\frac{1}{2\sqrt{x+1}}\right]_{x=0}+...[/tex]
 
  • #3
260
1
The linear approximation of [itex]f(n)[/itex] about [itex]n_0[/itex] is given by:

[tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]


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

[tex]L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1[/tex]
 
  • #4
107
0
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:

[tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]

Thanks,
Len


The linear approximation of [itex]f(n)[/itex] about [itex]n_0[/itex] is given by:

[tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]


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

[tex]L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1[/tex]
 
  • #5
757
0
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.
 
  • #6
1,998
282
Even simpler, since

[tex] n + 1 \approx n + 1 + \frac {n^2} {4} = (n+\frac{1}{2})^2 [/tex]

if n<<1


[tex] \sqrt {n+1} \approx n + \frac{1}{2} [/tex]
 
  • #8
260
1
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:

[tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]

Thanks,
Len
You use a tangent line to approximate the function at [itex]n_0[/itex]. The equation of the tangent line is obviously given by [itex]y=f'(x_0)(x-x_0)+f(x_0)[/itex].

http://www.function-grapher.com/graphs/fc544b3a572de466f056ae4620193c3f.png [Broken]
 
Last edited by a moderator:
Top