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Recurrence Relation T(n)=T(n/8)+T(n/4)+lg(n)

I have been doing recurrence relations, but I just can't even get this one started. I tried iteration method, but don't think that's a good way to go about it and couldn't get anywhere. A friend told me master method should work, and another said substitution, but haven't had luck with either.

Edit: Just started learning, so assume I'm a beginner to this material.

Edit2: I've been told the answer is $n\lg(n)$ (base-2), but I still can't see how you would arrive to that.

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    This should $a_n = T(2^n)$ help! – 2017-02-01
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    $a_n=T(2^n)$ in the last comment does help a bit, but that leads to a linear non-homogenous recurrence whose homogenous solution pathway involves using the cubic formula. I do not think there is an easier way about going about things. – 2017-02-01
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    The edit $2$ solution is definitely false. If lg means log_10 x or even ln x. – 2017-02-01
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    Do you want an exact solution, or do you want an asymptotic approximation? – 2017-02-01
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    @AhmedS.Attaalla base 2, edited. – 2017-02-01
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    @apnorton: I'd take anything at this point. If $nlg(n)$ is right, then a show of how to find the constant(s) c that would bound the relation to $\theta(nlg(n))$. If it's not, then asymptotic, since I need to find the asymptotic upper and lower bounds. Hell, even if it is right I'd appreciate how you'd find the asymptotic approximation. – 2017-02-01
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    "Just started learning" Then perusing some answers to questions on the subject on the site should be a start, otherwise we will get the nth installment of the ultra classical method applied to your case. – 2017-02-01
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    @Did I've been here long enough to know to look before I ask, but found nothing that seemed to actually fit with mine, or had answers that didn't really explain the process (though that was more often results of google search). I've been (quickly) taught iteration, substitution, and master method, but essentially every example was pretty and had pretty/neat solutions (calculations included). This is being asked after hours of working with others and getting nowhere, research included. If there's a simple or classical way for dealing with my case, I just don't see it. – 2017-02-01
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    Which part of the answer below was escaping you and your friends? – 2017-02-01
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    @Bob Sorry for interfering here. Your problem is "very" classical, indeed: it is a linear recursion relation with constant coefficients, and with an inhomogneity (n) (see Did's solution). Start with the homogeneous relation, make the ansatz a(n) = q^n. You find that q must obey q^3= 1+q. The leading term of a(n) for n->oo is determined by the root q with the greatest module, this is q1 = 1.32472 (a polynomial inhomogeneity is always weaker). Hence the leading term is a(n) ~ Exp( n Log(q1)) or Log(a(n))/n ~= 0.2812, the result of Did. – 2017-02-02

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If $a_n=T(2^n)$, then, for $n\geqslant3$, $$a_n=a_{n-3}+a_{n-2}+n$$ hence the series $$A(x)=\sum_{n=0}^\infty a_nx^n$$ solves $$A(x)=a_0+a_1x+a_2x^2+\sum_{n=3}^\infty(a_{n-3}+a_{n-2}+n)x^n$$ that is, $$A(x)=a_0+a_1x+a_2x^2+x^3A(x)+x^2(A(x)-a_0)+B(x)$$ with $$B(x)=\sum_{n=3}^\infty nx^n=\frac{x}{(1-x)^2}-x-2x^2$$ or equivalently, $$A(x)=\frac{C(x)}{1-x^2-x^3}$$ with $$C(x)=a_0+a_1x+(a_2-a_0)x^2+B(x)$$ Now, there exists some triplets of complex numbers $(u_k)$ and $(v_k)$ such that $$1-x^2-x^3=\prod_{k=1}^3(1-u_kx)$$ and $$\frac1{1-x^2-x^3}=\sum_{k=1}^3\frac{v_k}{1-u_kx}=\sum_{k=1}^3v_k\sum_{n=0}^\infty (u_kx)^n$$ hence $$A(x)=C(x)\sum_{n=0}^\infty\sum_{k=1}^3v_ku_k^nx^n$$ that is, for every $n\geqslant3$, $$a_n=\sum_{k=1}^3\left(a_0u_k^n+a_1u_k^{n-1}+(a_2-a_0)u_k^{n-2}+\sum_{j=3}^n ju_k^{n-j}\right)v_k$$ In particular, the radius of convergence $R$ of $A(x)$ is the zero of smallest modulus of $1-x^2-x^3$, that is, $$R\approx0.75488$$ and this implies that $$\frac{\log a_n}n\to-\log R\approx0.28120$$ In this sense, $$T(2^n)=a_n\approx R^{-n}\approx(1.3247)^n$$ hence the asymptotics $$T(n)\approx n\lg n$$ is not correct, actually one can suspect that $$T(n)\approx n^\alpha$$ with $$\alpha=-\log_2R\approx0.40568$$

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    (+1) Thank you for sharing this calculation with us although it is "classical" and so not really exciting for the experts here (but it is for us noobs). – 2017-02-03
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    @user144921 Thanks. "Not really exciting" is offtopic here, but the fact that tons of variants of this computation are already present on the site should indeed be relevant to "you noobs"... – 2017-02-03