Use linear approximation, i.e. the tangent line, to?
A linear approximation function is given by: L(x) = f(a) + f '(a)*(x-a) where a is the point where the linear approximation is at (in this case, 36) First, evaluate f(36) = sqrt(36) = 6 Next, take the derivative of f(x) f(x) = sqrt(x) f '(x) = 1/(2*sqrt(x) f '(36) = 1/(2*sqrt(36)) = 1/12 Then put these numbers back into the L(x) eqn and evaluate it at x=36.4. L(x) = 6 + (1/12)*(x-36) L(36.4) = 6 + (1/12)*(36.4-36) = 6 + 4/120 = 181/30 To specifically answer your question, m = 1/12 b = 6 L(36.4) = 181/30
Use linear approximation, i.e. the tangent line, to approximate ?
Use linear approximation, i.e. the tangent line, to approximate (sqrt(125.3))^(1/3) as follows: Let F(x)+ (sqrt(X))^(1/3). The equation of the tangent line to f(x) at x=125 can be written in the form y=mx+b where m is: ? and where b is: ? Using this, we find our approximation for (sqrt(125.3))^(1/3) is ? I need help finding the question marks.
Use linear approximation, i.e. the tangent line, to approximate 5.98^2 as follows: Let f(x) = x ^2.?
Use linear approximation, i.e. the tangent line, to approximate 5.98^2 as follows: Let f(x) = x ^2. The equation of the tangent line to f(x) at x = 6 can be written in the form y = mx+b where: m=? b=? Using this, we find our approximation for 5.98 ^2 is ?
Use linear approximation, i.e. the tangent line, to approximate 4.7^7 as follows: Let f(x) = x^7. The?
With f(x) = x^7, we have f'(x) = 7x^6. The slope of the linear approximation line at x = 5 is: f'(5) = 7(5)^6 = 109375. Then, since f(x) passes through (5, 5^7) ==> (5, 78125), the equation of the linear approximation line at x = 5 is: y - 78125 = 109375(x - 5) ==> y = 109375x - 468750. This gives: (4.7)^7 ≈ 109375(4.7) - 468750 ≈ 45312.5. I hope this helps!
How do you prove Newton-Raphson method?
I think the point of the question is not to “derive” the method, rather, to prove that it actually finds roots. It’s usually ok to say that if a root is given approximately by some computation, then iterating that computation “should” get us closer, but “should” isn’t satisfactory for a proof.We begin by expressing [math] f [/math] in its taylor series about an estimate [math] x_n [/math] of a root [math] x_r [/math], truncating at the third term. Note that this assumes our initial estimate is good enough, since we rely on the taylor series being a good approximation of the function.[math] f(x_r) = f(x_n + \delta) = f(x_n) + f’(x_n)(x_r - x_n) + \frac{f’’(x_n)}{2}(x_r - x_n)^2 + \mathcal{O}(\delta^3) [/math].Now in order to insert our next estimate [math] x_{n+1} [/math], rearrange the Newton Raphson scheme to get[math] f(x_n) = f’(x_n)(x_n - x_{n+1}) [/math]Substitute this into our taylor series to obtain[math] f(x_r) = f’(x_n)(x_n - x_{n+1}) + f’(x_n)(x_r - x_n) + \frac{f’’(x_n)}{2}(x_r - x_n)^2 + \mathcal{O}(\delta^3) [/math]Finally, using the fact that [math] x_r [/math] is a root, we can say [math] f(x_r) = 0 [/math]. Use this to set the left hand side to 0 and rearrange to get[math] 0 = f’(x_n)(x_r - x_{n+1}) + \frac{f’’(x_n)}{2}(x_r - x_n)^2 + \mathcal{O}(\delta^3) [/math].Now define the error in our estimates as [math] e_n = x_r - x_n [/math]. One final rearrangement to solve for [math] e_{n+1} [/math] gives us[math] e_{n+1} = -\frac{f’’(x_n)}{2f’(x_n)}e_n^2 +\mathcal{O}(\delta^3) [/math].We have absorbed a constant into the big O here. Note that in this rearrangement, we divided by [math] f’(x_n) [/math]. This only works if the derivative at a particular iteration is not 0. This last expression tells us that after each iteration, the error in our new estimate is proportional to the square of our old estimate (and even tells us what this constant is).This proof not only proves the Newton Raphson method converges, it tells us we need a good enough initial estimate, gives us an idea of how fast the convergence is, and gives us a condition that the derivative be non-zero about the root. This causes problems if the derivative of the function is zero at the root, but one can show this reduces to linear convergence, rather that quadratic.
If f(x) is a polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16. Find the value of f(5)?
Let us assume : [math]f(x) = ax^3 + bx^2 + cx + d [/math]Then we get, [math]f(1) = a + b + c + d = 1 ... (1)[/math][math]f(2) = 8a + 4b + 2c + d = 2 ... (2)[/math][math]f(3) = 27a + 9b + 3c + d = 3 ...(3)[/math][math]f (4) = 64a + 16b + 4c + d = 16 ...(4)[/math]Solving (1) and (2), We get[math]7a + 3b + c = 1 ...(5)[/math]Solving (1) and (3), We get[math]26a + 8b + 2c = 2 ...(6)[/math]Solving (1) and (4), We get[math]63a + 15b + 3c = 15 ...(7)[/math]Solving (5) and (6)[math]6a + b = 0[/math]Solving (5) and (7)[math]7a + b = 2[/math]This gives, a = 2, b = -12And then, c = 23 and d = - 12Thus,[math] f(x) = 2x^3 - 12x^2 + 23x - 12[/math]And So, [math]f(5) = 2*(5)^3 - 12*(5)^2 + 23(5) - 12 = 53[/math]The required answer is f(5) = 53