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**Mike****Guest**

I have this mathematics question that is really giving me a headache.......

PLEASE HELP!!!

Show that the equation 2x-tanx=0 has a root in the interval [1.1, 1.2]. hence find this root correct to 2 decimal places.

b) Find the Macluarin series expansion for f(x) = x2 + sin (x/2) up to and including the term in x5.

Please HELP........

Deeply appreciated.

Thank you

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

The first one is 1.165561185 and then my little solar calculator couldn't go any further. It took six iterations using Newton's Method.

x - f(x)/f '(x)

Oh, and be sure to use radians because degrees will not give a solution.

You can use degrees but then your equation will be: y = 2x - tan(180x/pi)

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

i was working on problem one in degree mode and my calculator kept spitting out positiive results even for very large values of x. then i switched over to radian mode and it worked! sometimes a lil trial and error will get you on the right track!

2)

f(x) = x² + sin(x/2) f(0) = 0

f'(x) = 2x + (1/2)cos(x/2) f'(0) = 1/2

f''(x) = 2 - (1/4)sin(x/2) f''(0) = 0

f'''(x) = -(1/8)cos(x/2) f'''(0) = -1/8

f''''(x) = (1/16)sin(x/2) f''''(0) = 0

f'''''(x) = (1/32)cos(x/2) f'''''(0) = 1/32

0 + [(1/2)x]/1! + 0x²/2! - [(1/8)x³]/3! + [0x^(4)]/4! + [(1/32)x^5]/5!

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

To show that there is a root between 1.1 and 1.2, just find the values of the functions at each of those points and show that there is a change of sign.

f(1.1) ≈ 0.23

f(1.2) ≈ -0.17

There is a change of sign, so a root must exist between the two.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Just don't forget to also state the function is continuous from [1.1, 1.2].

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

True. That method wouldn't work if the range had a number of the form (n+0.5)π in it, because the sign changes at every point with that form, but they are not roots.

Why did the vector cross the road?

It wanted to be normal.

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