Your question really depends on the values of x and y. And your expression doesn't have c. I assumed your function to be:
f(x,y)=\sqrt{a x^8+b x^4y^4+c y^8}
I only outline the method to obtain a power series here.
f(x,y)=c y^8\sqrt{1+\left(\frac{a x^8+b x^4y^4}{c y^8}\right)}
By...
I agree that differential has the meaning of limit already. But without saying so, it is rather difficult to explain why du dv = 0 in PhDorDust's example. Furthermore, I wonder if product of differentials must be zero, since we sometimes have dx dy in double integral. I think the reason is that...
Char. Limit, for the integration, what you have written is the standard way I learnt to do integration. The general form is:
\int g(x) f'(x)d x=\int g(x)d f(x)
For example, by using the property of differential, d x^2/2 = x d x
\int e^{x^2}x \text{dx}
=\frac{1}{2}\int...
For a more direct approach, you may try this:
f(x+\Delta x, y+\Delta y) = \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k} f(x,y+\Delta y)
= \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k}
\sum_{j=0}^{\infty} \frac{(\Delta y)^j}{j!} \frac{d^j}{dy^j}
f(x,y)...
You are right up to this step:
d u\cdot v = u \cdot dv + v \cdot du + du \cdot dv
However, if differential is small, differential multiplying to another differential is much smaller, which can be neglected (In limit sense). Therefore:
d u\cdot v = u \cdot dv + v \cdot du
Hello, alicexigao. I can't really help you prove the identity. But I think it maybe able to evaluate to something more simple.
\int _0^1\frac{d C(q(x)+k'(q'(x)-q(x)))}{d k'}d k'
=\int _0^1d C(q(x)+k'(q'(x)-q(x)))
=[ C(q(x)+k'(q'(x)-q(x)))]_0^1
=C(q(x)+(q'(x)-q(x)))-C(q(x))...
Dickfore is correct about the difficulties in finding inverse of infinity series. I have written a paper about finding power series of tan x + sec x. And I think Unit may take a look in it. For the secant series, it just corresponds to the even power of the series, as tan x is odd, and sec x is...
quasar987, negative logarithm does indeed exist.
http://en.wikipedia.org/wiki/Logarithm#Logarithm_of_a_negative_or_complex_number"
But I think you are right to assume x is positive.
Hello, quasar. I don't think using square root function is a good method, since you don't know whether to take the positive or negative square root. For more difficult case, you may not be able to find the inverse function.
To differentiate f(x) w.r.t g(x), just do the following:
\frac{d...
The key step that's incorrect should be you replace:
F_1=\delta (t-n T)\delta (\tau -n T)
with
F_2=\delta (t-\tau )
Clearly, F_1 is non-zero when both t = n T and \tau = n T. Though it is true that t = \tau, but you missed t = n T.
There is a second problem as well. A product of...
I think your original binomial series is wrong too.
The correct one should be summing about r, not n. i.e.
(1+x)^n = \sum_{r=0}^{\infty} \binom{n}{r} x^r
This formula is valid, even for complex n. You are right that this formula can be derived using Taylor series.
Can you give a full form of right-hand side? And are you sure your identity must be right? I get different result from yours.
\sum _{n=0}^{\infty } \frac{(-x)^nc(n)}{n!}=\sum _{k=0}^{\infty } \left(\sum _{n=0}^{\infty } \frac{(-x)^n(n-x)^k}{n!k!}\right)c^{(k)}(x)
For the first few terms...
1. Your equation is indeed correct. But I prefer to write it this way:
A_n=A_{n-1}y+A_1x
2. If you want to make it simple and compute it by hand, I guess you would like to solve the above difference equation. Here is the solution:
A_n=A_1y^{n-1}+A_1x\left(\frac{y^{n-1}-1}{y-1}\right)...