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  1. R

    Taylor Series in Multiple Variables

    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...
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    Wikipedia shows a proof of product rule using differentials 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...
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    Derivatives wrt functions

    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...
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    Derive Multivariable Taylor Series

    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)...
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    Wikipedia shows a proof of product rule using differentials by

    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
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    Functional derivative

    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))...
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    Secant series

    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...
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    How to differentiate y = 8 ln x - 9 x with respect to x^2?

    quasar987, negative logarithm does indeed exist." But I think you are right to assume x is positive.
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    How to differentiate y = 8 ln x - 9 x with respect to x^2?

    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...
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    Double integral of product of Diracs

    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...
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    Binomial series vs Binomial theorem, scratching my head for three days on this

    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.
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    Asymptotic formula for a power 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...
  13. R

    How Can I Make This Simpler?

    You are welcome. Is it some kind of homework problem? Or you encounter it in your self-study?
  14. R

    How Can I Make This Simpler?

    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)...