Can calculus solve this?

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.

Last edited:

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

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

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:

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

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

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

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:

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

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.

Even simpler, since

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

if n<<1

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

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

$$L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)$$
You use a tangent line to approximate the function at $n_0$. The equation of the tangent line is obviously given by $y=f'(x_0)(x-x_0)+f(x_0)$.