The main problem is that, Mathematica tries to solve your problem analytically first. So it plugs in symbolic x, and your function can't handle it.
Please refer to this question in http://www.voofie.com/concept/Mathematica/" [Broken]...
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...
Piano man. Here is a link in http://www.voofie.com/concept/Mathematics/" [Broken] that you maybe interested.
http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-numbers-from-power-series/" [Broken]
I derived the power series of the function sec x +...
Please substitute y = x^2 to your differential equation, and you will find that it is indeed one of the solution.
y = x^2
y' = 2x
y'' = 2
Therefore:
x^2 y'' -3x y' + 4y = 2 x^2 - 6 x^2 + 4x^2 = 0
Actually, you can find a close form for your summation expression. Please refer to http://www.voofie.com/concept/Mathematics/" [Broken] for details:
http://www.voofie.com/content/156/how-to-sum-sinn-q-and-cosn-q/" [Broken]
In short,
\sum _{n=-N}^N \cos (n \theta )=\cos (N \theta )+\cot...
Hello. psholtz. I thought you just do the integration without considering their dependence. It is interesting that x\sqrt{1-y^2} + y\sqrt{1-x^2} = C is a solution. Thank you for your information in the Jacobian as well.
In fact, what you have show is that, x^2 is one of the solution. Since for all other terms other than a_0, they have to be zero in order for the expression to equal to zero.
I am sorry, but your 2nd equation:
\sqrt{1-y^2}dx + \sqrt{1-x^2}dy = 0
cannot be integrated to get the 3rd equation:
x\sqrt{1-y^2} + y\sqrt{1-x^2} = C
You cannot integrate term by term, from dx to x, since \sqrt{1-y^2} depends on x, and \sqrt{1-x^2} depends on y.
In fact, if you...
Hello. Please refer to my article in http://www.voofie.com/concept/Mathematics/" [Broken]:
http://www.voofie.com/content/18/solving-system-of-first-order-linear-differential-equations-with-matrix-exponential-method/" [Broken]
What you really need is matrix exponential, instead of matrix...
For your first equation, please refer to this question in http://www.voofie.com/concept/Mathematics/" [Broken]:
http://www.voofie.com/content/152/how-to-prove-eat-e-at_0-eat-t_0/" [Broken]
I think you typed wrong in this formula:
exp(At)_t=0 = I
0 is not equal to I. And your what's your...
Hello ferry2, I have solved your 2nd equation here:
http://www.voofie.com/content/146/how-to-solve-2x12-y--2-2x1-y-4-y-0/" [Broken]
And the solution is given by:
y(x) = C_1(2x+1) +C_2 (2x+1) \ln (2x+1)
Generating Function of your number
soroush1358, I have answer your question in this post: http://www.voofie.com/content/144/sum-of-subsets-with-k-elements-having-a-total-sum-of-r/" [Broken].
You can read more Mathematics related articles in http://www.voofie.com/concept/Mathematics/"...
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...
You may just write out the recurrence equation and solve it yourself:
a_n = \frac{a_{n+1}+a_{n-1}}{2} +1
a_1=1
a_{12}=12
The solution is:
a_n = -n^2 + 14 n -12
Hello, beetle2. It didn't mentioned 2nd order with variable coefficients, since there is no general method in solving that. However, what you can do is by guessing the first solution. Then you can find the 2nd one using various method, such as...
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.
In this article in http://www.voofie.com/concept/Mathematics/" [Broken]:
http://www.voofie.com/content/133/repeated-sum-and-partial-difference-equation/" [Broken]
I first gave a standard method in solving the problem (probably the same approach as your theorem). Then I solve it using a...
If it is not a must that you have to use method of undetermined coefficients, you can have a look of the operator method. In this case, you don't have to care about what kind of non-homogeneous function you have. At least you can write the solution in integral form. Please refer to my tutorial...