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# Taylor's series question [Solved!]

### My question

my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p

1. Taylor Series

### What I've done so far

I think I need to differentiate it several times.

Here it is:

f(x) = x^(1/2)

f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))

f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))

f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))

But I'm stuck then

X

my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p
Relevant page

<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>

What I've done so far

I think I need to differentiate it several times.

Here it is:

f(x) = x^(1/2)

f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))

f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))

f'''(x) =  3/8 x^(-5/2) = 3/(8 x^(5/2))

But I'm stuck then

## Re: Taylor's series question

Hello Snowboy

Your problem is very similar to the example on the page you came from:

1. Taylor Series

Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.

Or is you problem with the square root approximation?

Regards

X

Hello Snowboy

Your problem is very similar to the example on the page you came from:

<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>

Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.

Or is you problem with the square root approximation?

Regards

## Re: Taylor's series question

So a=1, right?

f(1) = 1

f'(1) = 1/2

f''(1) = -1/4

f'''(1) = 3/8

Plugging it into the formula:

1 + 1/2 (x-1)  -1/4(x-1)^2  + 3/8(x-1)^3-...

How many steps do we have to do?

X

So a=1, right?

f(1) = 1

f'(1) = 1/2

f''(1) = -1/4

f'''(1) = 3/8

Plugging it into the formula:

1 + 1/2 (x-1)  -1/4(x-1)^2  + 3/8(x-1)^3-...

How many steps do we have to do?

## Re: Taylor's series question

The question says you need 4 d.p. accuracy.

So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.

X

The question says you need 4 d.p. accuracy.

So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.

## Re: Taylor's series question

It says "at 1.1".

Is this OK?

sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3

 = 1 + 0.05 - 0.0025 + 0.000375  = 1.047875

I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.

Maybe I should do another step to make sure.

f''''(x) = -15/16 x^(-7/2)

f''''(1) = -15/16

So the 5th term is -15/16(x-1)^4

Plugging in 1.1 gives -0.00009375

Adding this gives 1.047875-0.00009375 = 1.04778125

To 4 d.p. it's still 1.0478.

But checking on my calculator, I get sqrt(1.1) = 1.04880884817

To 4 d.p., that's 1.0488. Is my answer wrong?

X

It says "at 1.1".

Is this OK?

sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3

= 1 + 0.05 - 0.0025 + 0.000375  = 1.047875

I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.

Maybe I should do another step to make sure.

f''''(x) = -15/16 x^(-7/2)

f''''(1) = -15/16

So the 5th term is -15/16(x-1)^4

Plugging in 1.1 gives -0.00009375

Adding this gives 1.047875-0.00009375 = 1.04778125

To 4 d.p. it's still 1.0478.

But checking on my calculator, I get sqrt(1.1) = 1.04880884817

To 4 d.p., that's 1.0488. Is my answer wrong?

## Re: Taylor's series question

Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.

Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).

To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:

The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.

You can see it's not a very good approximation as we get away from x=1.

X

Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.

Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).

To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:

[graph]310,250;-0.3,2.5;-0.5,2,0.5,0.5;sqrt(x),1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3[/graph]

The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.

You can see it's not a very good approximation as we get away from x=1.

## Re: Taylor's series question

Thanks a lot!

X

Thanks a lot!