The document discusses logarithms. Logarithms are the inverse of exponential functions. If y = ax, then x = loga y. The natural logarithm (ln) is the logarithm with base e. Logarithm laws and properties are presented, including the laws of logarithms, evaluating logarithmic expressions, and examples.
This document discusses backpropagation in convolutional neural networks. It begins by explaining backpropagation for single neurons and multi-layer neural networks. It then discusses the specific operations involved in convolutional and pooling layers, and how backpropagation is applied to convolutional neural networks as a composite function with multiple differentiable operations. The key steps are decomposing the network into differentiable operations, propagating error signals backward using derivatives, and computing gradients to update weights.
The document discusses logarithms. Logarithms are the inverse of exponential functions. If y = ax, then x = loga y. The natural logarithm (ln) is the logarithm with base e. Logarithm laws and properties are presented, including the laws of logarithms, evaluating logarithmic expressions, and examples.
This document discusses backpropagation in convolutional neural networks. It begins by explaining backpropagation for single neurons and multi-layer neural networks. It then discusses the specific operations involved in convolutional and pooling layers, and how backpropagation is applied to convolutional neural networks as a composite function with multiple differentiable operations. The key steps are decomposing the network into differentiable operations, propagating error signals backward using derivatives, and computing gradients to update weights.
The document discusses techniques for improving ad click prediction, a crucial task for online advertising. It presents the FTRL-Proximal learning algorithm, which combines the accuracy of online gradient descent with the sparsity of regularized dual averaging. Memory saving techniques are also key, including probabilistic feature inclusion, subsampling training data, and encoding values with fewer bits. Evaluation shows the approaches improve AUC by 11.2% over baselines. The techniques aim to efficiently handle massive online advertising data at scale.
In this work, we introduce a new Markov operator associated with a digraph, which we refer to as a nonlinear Laplacian. Unlike previous Laplacians for digraphs, the nonlinear Laplacian does not rely on the stationary distribution of the random walk process and is well defined on digraphs that are not strongly connected. We show that the nonlinear Laplacian has nontrivial eigenvalues and give a Cheeger-like inequality, which relates the conductance of a digraph and the smallest non-zero eigenvalue of its nonlinear Laplacian. Finally, we apply the nonlinear Laplacian to the analysis of real-world networks and obtain encouraging results.
The document discusses linear regression and regularization. It introduces linear regression models using basis functions and describes optimization methods like gradient descent. It then covers regularization, discussing how adding constraints like lasso and ridge regression can improve generalization by reducing overfitting. The document presents evidence selection methods like Bayesian model averaging to integrate over models. Finally, it describes applying empirical Bayes methods to estimate hyperparameters like α and β by maximizing the marginal likelihood.
Effective Optimization Algorithms for Blind and Supervised Music Source Separation with Nonnegative Matrix Factorization
長倉研究奨励賞第三次審査，20分間の研究概要説明
内容は自身の学位論文の一部に相当
The document discusses techniques for improving ad click prediction, a crucial task for online advertising. It presents the FTRL-Proximal learning algorithm, which combines the accuracy of online gradient descent with the sparsity of regularized dual averaging. Memory saving techniques are also key, including probabilistic feature inclusion, subsampling training data, and encoding values with fewer bits. Evaluation shows the approaches improve AUC by 11.2% over baselines. The techniques aim to efficiently handle massive online advertising data at scale.
In this work, we introduce a new Markov operator associated with a digraph, which we refer to as a nonlinear Laplacian. Unlike previous Laplacians for digraphs, the nonlinear Laplacian does not rely on the stationary distribution of the random walk process and is well defined on digraphs that are not strongly connected. We show that the nonlinear Laplacian has nontrivial eigenvalues and give a Cheeger-like inequality, which relates the conductance of a digraph and the smallest non-zero eigenvalue of its nonlinear Laplacian. Finally, we apply the nonlinear Laplacian to the analysis of real-world networks and obtain encouraging results.
The document discusses linear regression and regularization. It introduces linear regression models using basis functions and describes optimization methods like gradient descent. It then covers regularization, discussing how adding constraints like lasso and ridge regression can improve generalization by reducing overfitting. The document presents evidence selection methods like Bayesian model averaging to integrate over models. Finally, it describes applying empirical Bayes methods to estimate hyperparameters like α and β by maximizing the marginal likelihood.
Effective Optimization Algorithms for Blind and Supervised Music Source Separation with Nonnegative Matrix Factorization
長倉研究奨励賞第三次審査，20分間の研究概要説明
内容は自身の学位論文の一部に相当
2. • Tanım: f : [ a , b] → Rtanımlı iki fonksiyon olsun.Eğer F(x) in
türevi f(x)veya diferansiyeli f(x).dx olan F(x) fonksiyonunun
belirsiz integrali denir ve ∫ f (x).dx gösterilir.
biçiminde = F( x ) + c
• eşitliğinde; işaretine,integral işareti,f(x) e
integrand(integral altındaki fonksiyon),f(x).dx e diferansiyel
∫ f (x).dx = F(x) + cf(x) in ilkel fonksiyonu ve C ye integral sabiti
çarpanı,F(x) e
∫
denir.
3.
4. 1.Bir belirsiz integralin türevi,integrali alınan fonksiyona
eşittir:
∫ ( '
)
f ( x ).dx = (F( x ) + C)' = f ( x )
2.Bir belirsiz integralin diferansiyeli,integral işaretinin
(∫ f ( x).dx )
altındaki ifadeye eşittir: ′
d = f ( x ).dx
3.Bir fonksiyonun diferansiyelinin belirsiz integrali,bu
fonksiyon ile bir C sabitini toplamına eşittir:
∫ d ( f ( x)) = f ( x) + c
5. ∫4 x
5
Örnek-1- .dxbelirsiz integralinin türevini bulunuz.
Çözüm : d
dx
(∫4 x 5
)
.dx = x 5
4
Örnek-2-
∫ d ( x 3 + x) belirsiz integralini bulunuz.
Çözüm :
∫d (x + x) = x + x + c
3 3
Örnek-3- belirsiz integralinin diferansiyelini
∫ x +1.dx
2
bulunuz.
Çözüm :
∫ x 2 +1.dx = x 2 +1.dx
6. 1
1. ∫ x dx =
n
x n +1 +c (n ≠ −1)
n +1
1
2. ∫e x .dx = e x +c 3. ∫ dx = ln x + c
x
4. ∫a x .dx = 1 a x +c (a > 0, a ≠ 1)
ln a
5. ∫ sin x.dx = − cos x + c 6. ∫ cos x.dx = sin x + c
∫
7. ∫ tan x. sec x.dx = sec x + c 8. cot x. cos ecx.dx = − cos ecx + c
7. 1
9. ∫ sec xdx = ∫ dx = ∫ (1 + tan 2 x)dx = tan x + c
2
cos 2 x
1
∫ cos ec xdx = ∫ 2 dx = ∫ (1 + cot 2 x)dx = − cot x + c
2
10. sin x
1
11. ∫1 + x 2 dx = arctan x +c
1
12. ∫ 1 −x 2
dx =arcsin x +c
8. Örnek-1-
∫x 5 dx belirsiz integralini bulunuz.
1 6
Çözüm: I =∫x dx = x +c
5
6
Örnek-2- ∫ (e 3 +e x ) dx belirsiz integralini bulunuz.
Çözüm: I = ∫ (e 3 + e x )dx = e 3 .x + e x + c
x5 + x 4 − 2x
Örnek-3-
∫ x5 dx belirsiz integralini bulunuz.
1 2 dx
Çözüm:
I = ∫ 1 + − 4 .dx = x + ln x − 2.∫ 4 = x + ln x − 2.∫ x − 4 dx
x x x
x −3 2
= x + ln x − 2. = x + ln x + 3 + c
−3 3x
9. 3−x
Örnek-4- ∫ 3 1 +
x
dx belirsiz integralini bulunuz.
x
x 1 3x
Çözüm: I = ∫ 3 + dx = + ln x + c
x ln 3
Örnek-5- ∫ tan 2 xdx belirsiz integralini bulunuz.
Çözüm: I =
∫ (1 + tan x − 1) dx = ∫ (1 + tan 2 x ) dx − ∫ dx = tan x − x + c
2
∫
Örnek-6- cot 4 xdx integralini hesaplayınız.
Çözüm: cot 4 xdx = cos 4 x 1 4 cos 4 x 1 (sin 4 x)1
∫ ∫ sin 4 x dx = 4 ∫ sin 4 x dx = 4 ∫ sin 4 x
1
= ln sin 4 x + c
4
10.
11. f ( g (x ) ) g ' ( x)dxİntegralinde u=g(x) ve u ' = g ' ( x)dx
∫
Dönüşümü yapılarak integral
∫ f ( x ) du getirilir.
haline
∫ ( x − 2 x + 3).( x − x).dx
Örnek-1- 4 2 3 integralini hesaplayınız
Çözüm:
u = x − 2x + 3
4 2
⇒ du = (4 x 3 − 4 x).dx
du
du = 4( x − x).dx
3
4
= ( x 3 − x ).dx
du 1 3 1 u4 1 4
I = ∫ u 3 = ∫ u .du = +c I = ( x − 2 x 2 + 3) 4 + c
4 4 4 4 16
12. ∫e
sin x
Örnek-2- . cos x.dx integralini hesaplayınız.
Çözüm: u = sinx du = cosx.dx
I = ∫ e .du = e + c
u u
x
Örnek-3-
∫1 + x 2 dx integralini hesaplayınız.
Çözüm: u = 1+ x 2 du = 2xdx ⇒ du =x.dx
2
du
I =∫ 2 = 1 ln u + c = 1 ln(1 + x 2 ) + c
u 2 2
13. ln x
Örnek-4-
∫ x
dx integralini hesaplayınız.
Çözüm: u = ln x 1
du = dx
x
3
1
u 2
I =∫ u du = ∫ u du =
2
+c
3
2
3
2
= (ln x ) 2 + c
3
14. dx
Örnek-5- ∫ e x + 1 dx integralini hesaplayınız.
Çözüm:
dx ex + 1− ex ex + 1 ex ex
I = ∫ x dx = ∫ dx = ∫ x dx − ∫ x dx = ∫ dx − ∫ x dx
e +1 e +1
x
e +1 e +1 e +1
ex
I2 = ∫ x dx u = e + 1 ⇒ du = e .dx
x x
e +1
du
I2 = ∫ = ln u + c
u
I = x − ln e +1 + c
x
15. x
e
Örnek-6- ∫ x
dx integralini hesaplayınız.
Çözüm: 1 1
u= x du = dx 2du = dx
2 x x
I = ∫ e u .2du = 2 ∫ e u du = 2e u + c I = 2e + c x
Örnek-7- ∫ sin x. cos x.dx integralini hesaplayınız.
Çözüm: u = sin x du = cos x.dx
2 2
u sin x
I = ∫ u.du = +c I= +c
2 2
16. Örnek-8- ∫ x 2 − 4 x ( x − 2).dx integralini hesaplayınız.
Çözüm: u = x2 − 4x du = (2 x − 4 x).dx = 2( x − 2).dx
du
= ( x − 2).dx
2
3
3
du 1 1u 1 7 2
I= u = ∫ u .du = +c = u +c
2 2 2 3 3
2
3
1 2
I = ( x − 4 x) 2 + c
3
17. arctan x
Örnek-9- ∫ dx integralini hesaplayınız.
1+ x 2
u = arctan x du = 1
Çözüm: dx
1+ x 2
2
u arctan 2 x
I = ∫ u.du = + c I= +c
2 2
e x + e− x
Örnek-10-
∫ e x − e− x dx integralini hesaplayınız.
−x −x
u = e −e x
du = (e + e )dx
x
du −x
I =∫
u
= ln u + c I = ln e − e + c
x
18. Örnek-11-
∫ (cot x − tan x)dx integralini hesaplayınız.
Çözüm:
∫ cot xdx − ∫ tan xdx
I1 I2
u = sin x t = cos x
du = cos x dt = − sin x.dx
I = ln u + ln t + c
I = ln sin x + ln cos x + c
19. sin 2 x
Örnek-12- ∫ 3 + cos2 x dx integralini hesaplayınız.
Çözüm: u = 3 + cos 2 x du = − 2 cos x sin x = − sin x
du
I = ∫− = − ln u + c I = − ln 3 + cos 2 x + c
u
Örnek-13- ∫ (tan 4 x + tan 2 x )dx integralini hesaplayınız.
Çözüm: I = ∫ tan x(tan x + 1)dx
2 2
u = tan x du = (1 + tan x)dx 2
3 3
u tan x
I = ∫ u du = + c
2
I= +c
3 3
20. dx
Örnek-14- ∫ 9 − 25 x 2
integralini hesaplayınız.
Çözüm:
dx 1 bx
a, b ∈ R − { 0} ⇒ ∫ a −b x
2 2 2
= arcsin + c
b a
dx 1 5x
∫ 9 − 25 x 2
= arcsin + c
5 3
21. sin 2 x = 2 sin x. cos x
1. sin 2 x + cos 2 x = 1 3.
2. sec 2 x − tan 2 x = 1 4. cos 2 x = 2. cos 2 x − 1
= 1− 2 sin 2 x
1
*
sin a. sin b = − [ cos(a + b) − cos(a − b)]
2
1
* sin a. cos b = [ sin( a + b) + sin( a − b)]
2
1
* cos a. cos b = [ cos(a + b) + cos(a − b)]
2
22. Örnek-1-
∫ cos 4 x.cos 2 x.dx integralini hesaplayınız.
Çözüm:
1 11 1
I = ∫ (cos 6 x + cos 2 x).dx = sin 6 x + sin 2 x + c
2 26 2
1 1
I = sin 6 x + sin 2 x + c
12 4
23. Örnek-2- ∫ sin 2 x.dx integralini hesaplayınız.
1 − cos 2 x 1 1
Çözüm: ∫ sin x.dx = ∫
2
dx = ∫ dx − ∫ cos 2 x.dx
2 2 2
1 1
= sin 6 x + sin 2 x + c
12 4
∫ cos
2
Örnek-3- xdx integralini hesaplayınız.
1 + cos 2 x 1 1
Çözüm: ∫ cos xdx = ∫ dx = ∫ dx + ∫ cos 2 xdx
2
2 2 2
x 1
I = + sin 2 x + c
2 4
24. Örnek-4- ∫ sin 4 x.dx integralini hesaplayınız.
Çözüm:
2
1 − cos 2 x 1
∫ sin x.dx = ∫ (sin x) .dx = ∫ 2 dx = 4 ∫ (1 − cos 2 x) .dx
4 2 2 2
1 1 1
= ∫ (1 − 2 cos 2 x + cos 2 x)dx = ( x − 2. sin 2 x + ∫ cos 2 xdx)
2 2
4 4 2
1 + cos 4 x
=
2
x 1 1 1 + cos 4 x x 1 1 1
I = − sin 2 x + ∫ dx = − sin 2 x + ( x + sin 4 x) + c
4 4 4 2 4 4 8 4
3 x sin 2 x 1
I= − + sin 4 x + c
8 4 32
25. 5
Örnek-5- sin xdx integralini hesaplayınız.
Çözüm:
sin xdx = ∫ (sin x) .sin x.dx = ∫ (1 − cos x) .sin x.dx
5 2 2 2 2
du = − sin xdx
u = cos x I = ∫ (1 − u ) .(− du )
2 2
− du = sin x.dx
I = ∫ (1 − 2u 2 + u 4 ).(−du ) = ∫ (−1 + 2u 2 − u 4 ).du
5
2 3 u 2 3 1 5
I = −u + u − I = − cos x + cos x − cos x + c
3 5 3 5
26. ∫ u.du = u.v - ∫ v.du
YARDIM:
1)dv’nin integrali kolay olmalı.
2) ∫ v.du integrali ilk integral
3) u’yu seçerken genelde aşağıdaki sıra ile seçmek avantajlıdır.
Logaritma Arc Polinom Trig. Üstel f.
29. ÖRNEK: x + 2x + x + 2
3 2
∫ x+1
.dx
x 3 + 2 x 2 + x + 2 =x2+x kalan:2
x+1
= x2 + x + 2
∫ 3 2x +1
.dx
= x + x + 2 ln x + 1 + c
3 2
30. x + 2x + 3
2
Örnek:
∫ x 3 − x .dx
x2 + 2x + 3 A B C
= + +
x( x − 1)( x + 1) x x − 1 x + 1
x + 2 x + 3 = A( x − 1) + Bx( x + 1) + Cx ( x − 1)
2 2
B=3 ; C=1 ;A=-3
−3 3 1
∫ x x − 1 x + 1 .dx
+ +
=-3ln|x|+3ln|x-1|+ln|x+1|+c
31. Sadece köklü ifade varsa!!!
a
* a −b x
2 2 2
=> x = sin u
b
a
* a +b x
2 2 2
=> x = tan u
b
a
* b x −a
2 2 2
=> x = sec u
b
32. dx
∫ 4 x 2 + 4 x + 17 = ???
4x +2
4 x + 17 =(2x+1)2+42
dx 4 x 2 + 4 x +17
∫ 4 x 2 + 4 x +17 = ( 2 x +1) +16
2
2 x +1 = 4 tan u
2 x = 4 tan u −1 = ( 4 tan u ) 2 +16
2dx = 4 + 4 tan u 2
= 4 tan 2 u +1
dx = 2 + 2 tan u 2
(
2 1 + tan 2 u
=
1 )
1 + tan u
2
(
dx = 2 1 + tan 2 u ) 4 1 + tan 2 u 2
DEVAMI
33. 1 1 1
= ∫ sec u = ∫ sec u = ln sec u + tan u + c
2
2 2 2
1 + tan u + tan u
2
1 4 x + 4 x + 17 2 x + 1
2
= ln 1 + +
2 16 4
1
= ln 4 x + 4 x + 17 + 2 x + 1 + c
2
2
34. ∫ arc sec x.dx = ???
dx
∫arc sec x.dx =x.arc sec x −∫ x2 −1
u =arc sec x
dx
du = = u =x; du =dx
>
x −
2
1
dx
=xarc sec x − ∫ x2 − x1
1 1 1
x =sec u = x =
> = cos u = cos u =−
> ln
cos u x x
1
xarc sec x − −
ln
+
c
x
1
xarc sec x +ln +c
x
xarc sec x − x +
ln c
35. b b
∫ f ( x ).dx = F ( x) = F (b) − F (a) c yok ; c-c=0
a a
36.
37. h( x)
F ( x) = ∫ f (u ).du ⇒
g ( x)
F ' ( x) = f (h( x).h' ( x) − f ( g ( x)).g ' ( x)
![• Tanım: f : [ a , b] → Rtanımlı iki fonksiyon olsun.Eğer F(x) in
türevi f(x)veya diferansiyeli f(x).dx olan F(x) fonksiyonunun
belirsiz integrali denir ve ∫ f (x).dx gösterilir.
biçiminde = F( x ) + c
• eşitliğinde; işaretine,integral işareti,f(x) e
integrand(integral altındaki fonksiyon),f(x).dx e diferansiyel
∫ f (x).dx = F(x) + cf(x) in ilkel fonksiyonu ve C ye integral sabiti
çarpanı,F(x) e
∫
denir.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)