- #1

mezarashi

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Seems like LaTeX isn't working -_-

(D^2 + r^2)^(1/2)

if r << D

( D^2( 1 + (r^2)/(D^2) ) )^(1/2)

D1 + (r^2)/(2D^2) <--- ???

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- Thread starter mezarashi
- Start date

- #1

mezarashi

Homework Helper

- 650

- 0

Seems like LaTeX isn't working -_-

(D^2 + r^2)^(1/2)

if r << D

( D^2( 1 + (r^2)/(D^2) ) )^(1/2)

D1 + (r^2)/(2D^2) <--- ???

.

.

.

- #2

- 11

- 1

Let's set this in Latex:

[tex](D^{2}+r^{2})^{\frac{1}{2}}=D(1+\epsilon)^{\frac{1}{2}}, \epsilon=(\frac{r}{D})^{2}[/tex]

If we now have [itex]\epsilon<<1[/itex], we have, by retaining the second term of the Taylor series about [itex]\epsilon=0[/itex]:

[tex](1+\epsilon)^{\frac{1}{2}}\approx{1}+\frac{\epsilon}{2}[/tex]

Thus, you have:

[tex](D^{2}+r^{2})^{\frac{1}{2}}\approx{D}+\frac{r^{2}}{2D}, \frac{r}{D}<<1[/tex]

[tex](D^{2}+r^{2})^{\frac{1}{2}}=D(1+\epsilon)^{\frac{1}{2}}, \epsilon=(\frac{r}{D})^{2}[/tex]

If we now have [itex]\epsilon<<1[/itex], we have, by retaining the second term of the Taylor series about [itex]\epsilon=0[/itex]:

[tex](1+\epsilon)^{\frac{1}{2}}\approx{1}+\frac{\epsilon}{2}[/tex]

Thus, you have:

[tex](D^{2}+r^{2})^{\frac{1}{2}}\approx{D}+\frac{r^{2}}{2D}, \frac{r}{D}<<1[/tex]

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- #3

mezarashi

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Hey awesome, thanks. Oh and Latex is working :P

Could you tell me a bit more about this "Taylor series" approximation or a link to its derivation, although I know this is true by punching some test numbers in the calculator. Thanks again.

Could you tell me a bit more about this "Taylor series" approximation or a link to its derivation, although I know this is true by punching some test numbers in the calculator. Thanks again.

Last edited:

- #4

- 11

- 1

Okay, about Taylor series:

You know that given f(x) differentiable, you may write it as:

[tex]f(x)=f(0)+\int_{0}^{x}f'(t)dt(1)[/tex]

Now, use partial integration on the integral in the following manner:

[tex]\int_{0}^{x}f'(t)dt=(t-x)f'(t)\mid_{t=0}^{t=x}-\int_{0}^{x}(t-x)f''(t)dt(2)[/tex]

We also have:

[tex](t-x)f'(t)\mid_{t=0}^{t=x}=0*f'(x)+xf'(0)=xf'(0)[/tex]

thus, (2) may be written as:

[tex]f(x)=f(0)+f'(0)x+\int_{0}^{x}(x-t)f''(t)dt(3)[/tex]

where I have drawn the minus sign underneath the integral sign.

We may now rewrite (3) by noting:

[tex]\int_{0}^{x}(x-t)f''(t)dt=\frac{x^{2}}{2!}f''(0)+\frac{1}{2!}\int_{0}^{x}(x-t)^{2}f'''(t)dt[/tex]

that is:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{1}{2!}\int_{0}^{x}(x-t)^{2}f'''(t)dt(4)[/tex]

The emerging series has the form, for an infinitely differentiable function:

[tex]f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}[/tex]

where [itex]n![/itex] is the factorial 1*2*3...*n (0!=1), and [tex]f^{(n)}[/tex] denotes the n'th derivative of f. (f is considered its own 0'th derivative).

That infinite series is called the Taylor series of f with respect to 0, a finite, truncated version of it is called a Taylor series approximation to f.

If we have the identity containing f(x) on the left-hand side and a finite sum and the integral on the right-hand side (for example our (4)), we call the integral "the remainder", or "error term".

You know that given f(x) differentiable, you may write it as:

[tex]f(x)=f(0)+\int_{0}^{x}f'(t)dt(1)[/tex]

Now, use partial integration on the integral in the following manner:

[tex]\int_{0}^{x}f'(t)dt=(t-x)f'(t)\mid_{t=0}^{t=x}-\int_{0}^{x}(t-x)f''(t)dt(2)[/tex]

We also have:

[tex](t-x)f'(t)\mid_{t=0}^{t=x}=0*f'(x)+xf'(0)=xf'(0)[/tex]

thus, (2) may be written as:

[tex]f(x)=f(0)+f'(0)x+\int_{0}^{x}(x-t)f''(t)dt(3)[/tex]

where I have drawn the minus sign underneath the integral sign.

We may now rewrite (3) by noting:

[tex]\int_{0}^{x}(x-t)f''(t)dt=\frac{x^{2}}{2!}f''(0)+\frac{1}{2!}\int_{0}^{x}(x-t)^{2}f'''(t)dt[/tex]

that is:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{1}{2!}\int_{0}^{x}(x-t)^{2}f'''(t)dt(4)[/tex]

The emerging series has the form, for an infinitely differentiable function:

[tex]f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}[/tex]

where [itex]n![/itex] is the factorial 1*2*3...*n (0!=1), and [tex]f^{(n)}[/tex] denotes the n'th derivative of f. (f is considered its own 0'th derivative).

That infinite series is called the Taylor series of f with respect to 0, a finite, truncated version of it is called a Taylor series approximation to f.

If we have the identity containing f(x) on the left-hand side and a finite sum and the integral on the right-hand side (for example our (4)), we call the integral "the remainder", or "error term".

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