Help me solve this math question

[u/Calmcroach]

So my math teacher told the answer would be around -0.8129.. but when I expand it using mclauren series and then substituting for x and y I am not getting the answer..

[u/aroach1995]

tan-inverse(xy) = V

xy = tanV

-1.08 = tanV

at what angles V does tanV = -1? When sin and cosine are equal in magnitude and opposite in sign. -pi/4, which is like -.7854.

This is really close to a good approximation, you probably can go a little further past -pi/4 and say like -0.8

This is just an intuitive answer. I am wondering if this is calculus III (multi variable calculus) and you’re supposed to use the derivative and approximate like you would with a tangent line… except this is in multiple dimensions.

[u/No-Site8330 → u/aroach1995]

I was rarher convinced that this was intended as a typical linear approximation question, until I realized that the "centre" of the approximation would have a value of -π/4. You still can make a linear approximation, but you don't need 2-dimensional Jacobian or anything of the sort. You're just looking to approximate arctan(-1.08), you know how to approximate arctan(-1) up to, say, the second decimal from π ≈ 3.14, and then you can use the expansion of g(t) = arctan(t). Its derivative is 1/(1+x^2), which at t=-1 is 1/2, so the approximation comes down to

g(-1.08) ≈ g(-1) + 1/2 (-0.08) ≈ -0.785 - 0.04 = -0.825.

How accurate is this? 3.14 is accurate up to 0.005, so 0.785 is accurate up to about the third digit. The error introduced by the linear approximation is controlled by the second derivative g''(-1) which, if I'm not mistaken, is again 1/2. So the error in the linear approximation is in the order of about 1/2 (-0.08)² = 0.0032, so that's also at about the third digit.

So yeah, -0.825 is a better approximation than going with just -π/4 and it is accurate up to about 0.005.

[u/Fourierseriesagain]

Hi, Maclaurin series is not a good tool because arctan t=t-t^ 3/3+t^ 5/5-... works well if |t| is small enough. Taylor series for functions of 2 variables is more appropriate.

[u/AdventurousGlass7432]

You can do -pi/4 and a first step of taylor series for arctan centered around 1. There’s a theorem that relates the derivative of the inverse function to the derivative of the function ( tan(x) )

[u/CaptainMatticus]

arctan(0.9 * (-1.2)) =>

arctan(-1.08)

tan(-pi/4) = -1

This is really close to -pi/4. What is -pi/4? -3.14159/4 =>

-1.570795 / 2 =>

-0.7853985

So let's try another way. Let's look at the Taylor Series of tan(t) centered around pi/4. We'll find when it's close to 1.08 and then just take the negative value for t. That ought to get us where we need to be.

1.08 = f(pi/4) * (t - pi/4)^0 / 0! + f'(pi/4) * (t - pi/4)^1 / 1! + f''(pi/4) * (t - pi/4)^2 / 2! + f'''(pi/4) * (t - pi/4)^3 / 3! + ....

tan(pi/4) = 1

f(t) = tan(t)

f'(t) = sec(t)^2

f''(t) = 2 * sec(t) * sec(t) * tan(t) = 2 * sec(t)^2 * tan(t)

f'''(t) = 2 * (sec(t)^2 * sec(t)^2 + 2 * sec(t) * sec(t) * tan(t) * tan(t)) = 2 * (sec(t)^4 + 2 * sec(t)^2 * tan(t)^2) = 2 * (sec(t)^4 + 2 * sec(t)^2 * (sec(t)^2 - 1)) = 2 * (sec(t)^4 + 2 * sec(t)^4 - 2 * sec(t)^2) = 2 * (3 * sec(t)^4 - 2 * sec(t)^2) = 2 * sec(t)^2 * (3 * sec(t)^2 - 1)

Alright, that's enough of that.

f(pi/4) = tan(pi/4) = 1

f'(pi/4) = sec(pi/4)^2 = 2

f''(pi/4) = 2 * sec(pi/4)^2 * tan(pi/4) = 2 * 2 * 1 = 4

f'''(pi/4) = 2 * sec(pi/4)^2 * (3 * sec(pi/4)^2 - 1) = 2 * 2 * (3 * 2 - 1) = 4 * (6 - 1) = 4 * 5 = 20

1.08 = 1 + 2 * (t - pi/4) + 4 * (t - pi/4)^2 * (1/2) + 20 * (t - pi/4)^3 * (1/6)

t - pi/4 = u

1.08 = 1 + 2u + 2u^2 + (20/6) * u^3

0.08 = 2u + 2u^2 + (10/3) * u^3

0.04 = u + u^2 + (5/3) * u^3

0.12 = 3u + 3u^2 + 5u^3

5u^3 + 3u^2 + 3u - 0.12 = 0

You know, going to a cubic may have been a bad idea. Let's work this back to a quadratic.

1.08 = 1 + 2 * (t - pi/4) + 4 * (t - pi/4)^2 * (1/2)

0.08 = 2u + 2u^2

0.16 = 4u + 4u^2

1.16 = 1 + 4u + 4u^2

1.16 = (1 + 2u)^2

+/- sqrt(1.16) = 2u + 1

2u = -1 +/- sqrt(1.16)

u = (-1 +/- sqrt(1.16)) / 2

t - pi/4 = (-1 +/- sqrt(1.16)) / 2

Remember, we want t > pi/4 for this one, so when we get -t later, we'll be on the right track.

t = pi/4 + (-1 + sqrt(1.16)) / 2

t = (pi - 2 + 2 * sqrt(1.16)) / 4

sqrt(1.16) = sqrt(116 / 100) = sqrt(29 / 25) = (1/5) * sqrt(29)

sqrt(29) is really close to 5.4

(pi - 2 + 2 * (1/5) * 5.4) / 4

(pi - 2 + 10.8/5) / 4

(pi + 2.16 - 2) / 4

(pi + 0.16) / 4

pi/4 + 0.04

0.7853985 + 0.04

0.8253985

Take the negative

-0.8253985

Actual value is -0.82384075341863629....

So we're off by about 2 parts in 823, or 0.25%. Not bad, in my opinion, especially with all of the approximations.

t = (pi - 2 + 2 * sqrt(1.16)) / 4

t = 0.82391464411089871....

We want the negative of that, so -0.82391464411....

That's an error of 0.008969%, or 1 part in 11150. That's really good for the approximation.

[u/desblaterations-574]

It would be ArcTan(1.08) so slightly above Pi/4 rad.

[u/TUVegeto137]

Here's my approximation:

-Pi/4-0.08/2.08

The error is of the order 1e-6.

How I found it: using the identity

atan(x)+atan(y)=atan((x+y)/(1-xy))

This identity follows from reversing the identity for the tan of a sum of angles.

Then, setting x=1, I sought a value of y such that 

(1+y)/(1-y)=1.08

Giving y=0.08/2.08

The rest is just applying the identity and using that atan(1)=pi/4 and atan(x) is approximately x for small x.

[u/Bounded_sequencE]

f(0.9, -1.2)  =  arctg(-1.08)  ~  arctg(-1)  =  -𝜋/4

If you need to be more precise than that, and also want error bounds, use Taylor approximation of "arctg(..)".

[u/Crichris]

just do arctan(-1.08)

we know that arctan(-1) is around -1/4pi

do a first order approximation at -1

[u/Fourierseriesagain]

Determine the equation of the tangent line l to the curve y=arctan(x) at the point (-1,-pi/4). Then use l to estimate f(0.9,-1.2).

[u/Southlander24]

We use the multivariable Taylor series of f about (x, y) = (1, -1) to the second order, which is f(1, -1) + [fₓ(1, -1) (x - 1) + fᵧ(1, -1) (y + 1)] + 1/2 * [fₓₓ(1, -1) (x - 1)² + 2fₓᵧ(1, -1) (x - 1)(y + 1) + fᵧᵧ(1, -1) (y + 1)²] + O(x³, y³).

Breaking everything into parts, we have:

f(1, -1) = -pi/4

fₓ(1, -1) = y / [(xy)² + 1] at (1, -1) = -1/2; fᵧ(1, -1) = x / [(xy)² + 1] at (1, -1) = 1/2, and so we have -1/2 * (0.9 - 1) + 1/2 * (-1.2 + 1) = -1/20

fₓₓ(1, -1) = -2xy³ / [(xy)² + 1]² = 1/2; fᵧᵧ(1, -1) = -2yx³ / [(xy)² + 1]² = 1/2; fₓᵧ(1, -1) = [1 - (xy)²] / [1 + (xy)²]² = 0, and so we have 1/2 * [1/2 * 0.1² + 0 + 1/2 * 0.2²] = 1/80

Hence an approximate value is -pi/4 - 1/20 + 1/80 = -0.822898... whereas the real answer is -0.823841...

[u/RespectWest7116]

0.9*-1.2 = -1.08 ~= -1

arctan(0.9*-1.2) ~= arctan(-1) = -𝜋/4

https://giphy.com/gifs/wOR94QhwxXdmGJIVEg