Your notation is a bit confusing. Normally an index "n" refers to integer values. What is it here? In other words, what does y2,0.1mean? Also there is no equation. What is that expression supposed to be equal to? And, finally, what is "t" here?
#3
leopard
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0
x'' - 2x' + x = 4t, x(0) = 1, x'(0)=1
Introducing y_1 = x and y_2 = x'
we have the system
y_1 ' = y_2
y_2 ' = 2y_2 - y_1 + 4t
right?
Using Euler's method with step size h=0.1, we get
y_{1, n+1} = y_{1, n} + 0.1y_{2,n}
y_{2,n + 1} + 0.1(2y_{2,n} - y_{1,n} + 4t)
It's easy to see that y_{1, 1} = 1 + 0.1 \cdot 1 = 1.1, but what is y_{2,1}?
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$
Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$
The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$
Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...